+ 
-
-
-
-> "_You help the developer community practice for interviews, and there is nothing better we could ask for._" -- [Alan Yessenbayev](https://opencollective.com/alan-yessenbayev)
-
## 版权
本项目著作权归 [GitHub 开源社区 Doocs](https://github.com/doocs) 所有，商业转载请联系 @yanglbme 获得授权，非商业转载请注明出处。
-## 联系我们
+## 联系我们 & 支持项目
欢迎各位小伙伴们添加 @yanglbme 的个人微信（微信号：YLB0109），备注 「**leetcode**」。后续我们会创建算法、技术相关的交流群，大家一起交流学习，分享经验，共同进步。
-| 
|
-| ------------------------------------------------------------------------------------------------------------------------------ |
+如果你觉得这个项目对你有帮助，也欢迎通过微信扫码赞赏我们 ☕️～
+
+| | 
|
+| -------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------- |
## 许可证
diff --git a/README_EN.md b/README_EN.md
index 002039e552df8..3b461059720a9 100644
--- a/README_EN.md
+++ b/README_EN.md
@@ -9,7 +9,8 @@
+ 
-
-
-
-> "_You help the developer community practice for interviews, and there is nothing better we could ask for._" -- [Alan Yessenbayev](https://opencollective.com/alan-yessenbayev)
-
## Copyright
The copyright of this project belongs to [Doocs](https://github.com/doocs) community. For commercial reprints, please contact [@yanglbme](mailto:contact@yanglibin.info) for authorization. For non-commercial reprints, please indicate the source.
+## Support Us
+
+If you find this project helpful, consider supporting us by buying us a coffee ☕️
+👉 [https://paypal.me/yanglbme](https://paypal.me/yanglbme)
+
## Contact Us
We welcome everyone to add @yanglbme's personal WeChat (WeChat ID: YLB0109), with the note "leetcode". In the future, we will create algorithm and technology related discussion groups, where we can learn and share experiences together, and make progress together.
diff --git a/images/pr-en.svg b/images/pr-en.svg
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diff --git a/images/pr.svg b/images/pr.svg
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+++ b/images/pr.svg
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diff --git a/images/starcharts.svg b/images/starcharts.svg
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\ No newline at end of file
diff --git a/images/support1.jpg b/images/support1.jpg
new file mode 100644
index 0000000000000..5d476dd5a74c4
Binary files /dev/null and b/images/support1.jpg differ
diff --git a/requirements.txt b/requirements.txt
index a861898c020f6..b87c0a5b8c9e6 100644
--- a/requirements.txt
+++ b/requirements.txt
@@ -1,2 +1,2 @@
black==24.3.0
-Requests==2.32.0
\ No newline at end of file
+Requests==2.32.4
\ No newline at end of file
diff --git a/solution/0000-0099/0001.Two Sum/README.md b/solution/0000-0099/0001.Two Sum/README.md
index f9a38ce75e350..88c20df42f7fa 100644
--- a/solution/0000-0099/0001.Two Sum/README.md
+++ b/solution/0000-0099/0001.Two Sum/README.md
@@ -353,6 +353,55 @@ class Solution {
}
```
+#### C
+
+```c
+#include 给你两个字符串 s 和 t ，统计并返回在 s 的 子序列 中 t 出现的个数，结果需要对 109 + 7 取模。
给你两个字符串 s 和 t ，统计并返回在 s 的 子序列 中 t 出现的个数。
测试用例保证结果在 32 位有符号整数范围内。
diff --git a/solution/0100-0199/0134.Gas Station/README.md b/solution/0100-0199/0134.Gas Station/README.md index 3ae6856d3f254..b350fca1ed379 100644 --- a/solution/0100-0199/0134.Gas Station/README.md +++ b/solution/0100-0199/0134.Gas Station/README.md @@ -57,10 +57,10 @@ tags:
提示:
gas.length == ncost.length == nn == gas.length == cost.length1 <= n <= 1050 <= gas[i], cost[i] <= 104n == gas.length == cost.length1 <= n <= 1050 <= gas[i], cost[i] <= 104-
查询 Employee 表中第 n 高的工资。如果没有第 n 个最高工资，查询结果应该为 null 。
编写一个解决方案查询 Employee 表中第 n 高的 不同 工资。如果少于 n 个不同工资，查询结果应该为 null 。
查询结果格式如下所示。
diff --git a/solution/0100-0199/0177.Nth Highest Salary/README_EN.md b/solution/0100-0199/0177.Nth Highest Salary/README_EN.md index 37c056f9d75ef..3373a1cd68980 100644 --- a/solution/0100-0199/0177.Nth Highest Salary/README_EN.md +++ b/solution/0100-0199/0177.Nth Highest Salary/README_EN.md @@ -31,7 +31,7 @@ Each row of this table contains information about the salary of an employee.-
Write a solution to find the nth highest salary from the Employee table. If there is no nth highest salary, return null.
Write a solution to find the nth highest distinct salary from the Employee table. If there are less than n distinct salaries, return null.
The result format is in the following example.
diff --git a/solution/0200-0299/0245.Shortest Word Distance III/README.md b/solution/0200-0299/0245.Shortest Word Distance III/README.md index 51d9475fc32ee..379385980b016 100644 --- a/solution/0200-0299/0245.Shortest Word Distance III/README.md +++ b/solution/0200-0299/0245.Shortest Word Distance III/README.md @@ -56,13 +56,12 @@ tags: ### 方法一：分情况讨论 -先判断 `word1` 和 `word2` 是否相等： +我们首先判断 $\textit{word1}$ 和 $\textit{word2}$ 是否相等： -如果相等，遍历数组 `wordsDict`，找到两个 `word1` 的下标 $i$ 和 $j$，求 $i-j$ 的最小值。 +- 如果相等，遍历数组 $\textit{wordsDict}$，找到两个 $\textit{word1}$ 的下标 $i$ 和 $j$，求 $i-j$ 的最小值。 +- 如果不相等，遍历数组 $\textit{wordsDict}$，找到 $\textit{word1}$ 和 $\textit{word2}$ 的下标 $i$ 和 $j$，求 $i-j$ 的最小值。 -如果不相等，遍历数组 `wordsDict`，找到 `word1` 和 `word2` 的下标 $i$ 和 $j$，求 $i-j$ 的最小值。 - -时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `wordsDict` 的长度。 +时间复杂度 $O(n)$，其中 $n$ 为数组 $\textit{wordsDict}$ 的长度。空间复杂度 $O(1)$。 @@ -199,6 +198,40 @@ func abs(x int) int { } ``` +#### TypeScript + +```ts +function shortestWordDistance(wordsDict: string[], word1: string, word2: string): number { + let ans = wordsDict.length; + if (word1 === word2) { + let j = -1; + for (let i = 0; i < wordsDict.length; i++) { + if (wordsDict[i] === word1) { + if (j !== -1) { + ans = Math.min(ans, i - j); + } + j = i; + } + } + } else { + let i = -1, + j = -1; + for (let k = 0; k < wordsDict.length; k++) { + if (wordsDict[k] === word1) { + i = k; + } + if (wordsDict[k] === word2) { + j = k; + } + if (i !== -1 && j !== -1) { + ans = Math.min(ans, Math.abs(i - j)); + } + } + } + return ans; +} +``` + diff --git a/solution/0200-0299/0245.Shortest Word Distance III/README_EN.md b/solution/0200-0299/0245.Shortest Word Distance III/README_EN.md index 72ab891b39ce6..6fbcb289984fb 100644 --- a/solution/0200-0299/0245.Shortest Word Distance III/README_EN.md +++ b/solution/0200-0299/0245.Shortest Word Distance III/README_EN.md @@ -45,7 +45,14 @@ tags: -### Solution 1 +### Solution 1: Case Analysis + +First, we check whether $\textit{word1}$ and $\textit{word2}$ are equal: + +- If they are equal, iterate through the array $\textit{wordsDict}$ to find two indices $i$ and $j$ of $\textit{word1}$, and compute the minimum value of $i-j$. +- If they are not equal, iterate through the array $\textit{wordsDict}$ to find the indices $i$ of $\textit{word1}$ and $j$ of $\textit{word2}$, and compute the minimum value of $i-j$. + +The time complexity is $O(n)$, where $n$ is the length of the array $\textit{wordsDict}$. The space complexity is $O(1)$. @@ -182,6 +189,40 @@ func abs(x int) int { } ``` +#### TypeScript + +```ts +function shortestWordDistance(wordsDict: string[], word1: string, word2: string): number { + let ans = wordsDict.length; + if (word1 === word2) { + let j = -1; + for (let i = 0; i < wordsDict.length; i++) { + if (wordsDict[i] === word1) { + if (j !== -1) { + ans = Math.min(ans, i - j); + } + j = i; + } + } + } else { + let i = -1, + j = -1; + for (let k = 0; k < wordsDict.length; k++) { + if (wordsDict[k] === word1) { + i = k; + } + if (wordsDict[k] === word2) { + j = k; + } + if (i !== -1 && j !== -1) { + ans = Math.min(ans, Math.abs(i - j)); + } + } + } + return ans; +} +``` + diff --git a/solution/0200-0299/0245.Shortest Word Distance III/Solution.ts b/solution/0200-0299/0245.Shortest Word Distance III/Solution.ts new file mode 100644 index 0000000000000..acb406b3b5ec7 --- /dev/null +++ b/solution/0200-0299/0245.Shortest Word Distance III/Solution.ts @@ -0,0 +1,29 @@ +function shortestWordDistance(wordsDict: string[], word1: string, word2: string): number { + let ans = wordsDict.length; + if (word1 === word2) { + let j = -1; + for (let i = 0; i < wordsDict.length; i++) { + if (wordsDict[i] === word1) { + if (j !== -1) { + ans = Math.min(ans, i - j); + } + j = i; + } + } + } else { + let i = -1, + j = -1; + for (let k = 0; k < wordsDict.length; k++) { + if (wordsDict[k] === word1) { + i = k; + } + if (wordsDict[k] === word2) { + j = k; + } + if (i !== -1 && j !== -1) { + ans = Math.min(ans, Math.abs(i - j)); + } + } + } + return ans; +} diff --git a/solution/0200-0299/0262.Trips and Users/README.md b/solution/0200-0299/0262.Trips and Users/README.md index 46938213c4cc8..87e992577500e 100644 --- a/solution/0200-0299/0262.Trips and Users/README.md +++ b/solution/0200-0299/0262.Trips and Users/README.md @@ -36,8 +36,6 @@ id 是这张表的主键（具有唯一值的列）。 status 是一个表示行程状态的枚举类型，枚举成员为(‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’) 。 --
表：Users
-
取消率 的计算方式如下：(被司机或乘客取消的非禁止用户生成的订单数量) / (非禁止用户生成的订单总数)。
编写解决方案找出 "2013-10-01" 至 "2013-10-03" 期间有 至少 一次行程的非禁止用户（乘客和司机都必须未被禁止）的 取消率。非禁止用户即 banned 为 No 的用户，禁止用户即 banned 为 Yes 的用户。其中取消率 Cancellation Rate 需要四舍五入保留 两位小数 。
+
Table: Users
+
The cancellation rate is computed by dividing the number of canceled (by client or driver) requests with unbanned users by the total number of requests with unbanned users on that day.
@@ -59,7 +59,7 @@ banned is an ENUM (category) type of ('Yes', 'No').Return the result table in any order.
-The result format is in the following example.
+The result format is in the following example.
Example 1:
diff --git a/solution/0200-0299/0269.Alien Dictionary/README.md b/solution/0200-0299/0269.Alien Dictionary/README.md index b1185f65b06a0..308a96369b8e8 100644 --- a/solution/0200-0299/0269.Alien Dictionary/README.md +++ b/solution/0200-0299/0269.Alien Dictionary/README.md @@ -288,6 +288,87 @@ public: }; ``` +#### Go + +```go +func alienOrder(words []string) string { + g := [26][26]bool{} + s := [26]bool{} + cnt := 0 + n := len(words) + for i := 0; i < n-1; i++ { + for _, c := range words[i] { + if cnt == 26 { + break + } + c -= 'a' + if !s[c] { + cnt++ + s[c] = true + } + } + m := len(words[i]) + for j := 0; j < m; j++ { + if j >= len(words[i+1]) { + return "" + } + c1, c2 := words[i][j]-'a', words[i+1][j]-'a' + if c1 == c2 { + continue + } + if g[c2][c1] { + return "" + } + g[c1][c2] = true + break + } + } + for _, c := range words[n-1] { + if cnt == 26 { + break + } + c -= 'a' + if !s[c] { + cnt++ + s[c] = true + } + } + + inDegree := [26]int{} + for _, out := range g { + for i, v := range out { + if v { + inDegree[i]++ + } + } + } + q := []int{} + for i, in := range inDegree { + if in == 0 && s[i] { + q = append(q, i) + } + } + ans := "" + for len(q) > 0 { + t := q[0] + q = q[1:] + ans += string(t + 'a') + for i, v := range g[t] { + if v { + inDegree[i]-- + if inDegree[i] == 0 && s[i] { + q = append(q, i) + } + } + } + } + if len(ans) < cnt { + return "" + } + return ans +} +``` + diff --git a/solution/0200-0299/0269.Alien Dictionary/README_EN.md b/solution/0200-0299/0269.Alien Dictionary/README_EN.md index 9677cf9f30970..24d8925dfb9a6 100644 --- a/solution/0200-0299/0269.Alien Dictionary/README_EN.md +++ b/solution/0200-0299/0269.Alien Dictionary/README_EN.md @@ -269,6 +269,87 @@ public: }; ``` +#### Go + +```go +func alienOrder(words []string) string { + g := [26][26]bool{} + s := [26]bool{} + cnt := 0 + n := len(words) + for i := 0; i < n-1; i++ { + for _, c := range words[i] { + if cnt == 26 { + break + } + c -= 'a' + if !s[c] { + cnt++ + s[c] = true + } + } + m := len(words[i]) + for j := 0; j < m; j++ { + if j >= len(words[i+1]) { + return "" + } + c1, c2 := words[i][j]-'a', words[i+1][j]-'a' + if c1 == c2 { + continue + } + if g[c2][c1] { + return "" + } + g[c1][c2] = true + break + } + } + for _, c := range words[n-1] { + if cnt == 26 { + break + } + c -= 'a' + if !s[c] { + cnt++ + s[c] = true + } + } + + inDegree := [26]int{} + for _, out := range g { + for i, v := range out { + if v { + inDegree[i]++ + } + } + } + q := []int{} + for i, in := range inDegree { + if in == 0 && s[i] { + q = append(q, i) + } + } + ans := "" + for len(q) > 0 { + t := q[0] + q = q[1:] + ans += string(t + 'a') + for i, v := range g[t] { + if v { + inDegree[i]-- + if inDegree[i] == 0 && s[i] { + q = append(q, i) + } + } + } + } + if len(ans) < cnt { + return "" + } + return ans +} +``` + diff --git a/solution/0200-0299/0269.Alien Dictionary/Solution.go b/solution/0200-0299/0269.Alien Dictionary/Solution.go new file mode 100644 index 0000000000000..b49abee4bad20 --- /dev/null +++ b/solution/0200-0299/0269.Alien Dictionary/Solution.go @@ -0,0 +1,76 @@ +func alienOrder(words []string) string { + g := [26][26]bool{} + s := [26]bool{} + cnt := 0 + n := len(words) + for i := 0; i < n-1; i++ { + for _, c := range words[i] { + if cnt == 26 { + break + } + c -= 'a' + if !s[c] { + cnt++ + s[c] = true + } + } + m := len(words[i]) + for j := 0; j < m; j++ { + if j >= len(words[i+1]) { + return "" + } + c1, c2 := words[i][j]-'a', words[i+1][j]-'a' + if c1 == c2 { + continue + } + if g[c2][c1] { + return "" + } + g[c1][c2] = true + break + } + } + for _, c := range words[n-1] { + if cnt == 26 { + break + } + c -= 'a' + if !s[c] { + cnt++ + s[c] = true + } + } + + inDegree := [26]int{} + for _, out := range g { + for i, v := range out { + if v { + inDegree[i]++ + } + } + } + q := []int{} + for i, in := range inDegree { + if in == 0 && s[i] { + q = append(q, i) + } + } + ans := "" + for len(q) > 0 { + t := q[0] + q = q[1:] + ans += string(t + 'a') + for i, v := range g[t] { + if v { + inDegree[i]-- + if inDegree[i] == 0 && s[i] { + q = append(q, i) + } + } + } + } + if len(ans) < cnt { + return "" + } + return ans +} diff --git a/solution/0200-0299/0275.H-Index II/README.md b/solution/0200-0299/0275.H-Index II/README.md index 546a3242358e5..47b3a5e04dfb9 100644 --- a/solution/0200-0299/0275.H-Index II/README.md +++ b/solution/0200-0299/0275.H-Index II/README.md @@ -17,7 +17,7 @@ tags: -给你一个整数数组 citations ，其中 citations[i] 表示研究者的第 i 篇论文被引用的次数，citations 已经按照 升序排列 。计算并返回该研究者的 h 指数。
给你一个整数数组 citations ，其中 citations[i] 表示研究者的第 i 篇论文被引用的次数，citations 已经按照 非降序排列 。计算并返回该研究者的 h 指数。
h 指数的定义：h 代表“高引用次数”（high citations），一名科研人员的 h 指数是指他（她）的 （n 篇论文中）至少 有 h 篇论文分别被引用了至少 h 次。
Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in ascending order, return the researcher's h-index.
Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in non-descending order, return the researcher's h-index.
According to the definition of h-index on Wikipedia: The h-index is defined as the maximum value of h such that the given researcher has published at least h papers that have each been cited at least h times.
总旅行距离 是朋友们家到聚会地点的距离之和。
-使用 曼哈顿距离 计算距离，其中距离 (p1, p2) = |p2.x - p1.x | + | p2.y - p1.y | 。
示例 1：
diff --git a/solution/0300-0399/0317.Shortest Distance from All Buildings/README_EN.md b/solution/0300-0399/0317.Shortest Distance from All Buildings/README_EN.md index 816bbabef6c3d..edd540c52e03d 100644 --- a/solution/0300-0399/0317.Shortest Distance from All Buildings/README_EN.md +++ b/solution/0300-0399/0317.Shortest Distance from All Buildings/README_EN.md @@ -32,8 +32,6 @@ tags:The total travel distance is the sum of the distances between the houses of the friends and the meeting point.
-The distance is calculated using Manhattan Distance, where distance(p1, p2) = |p2.x - p1.x| + |p2.y - p1.y|.
Example 1:
diff --git a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md
index 79d1666a074ef..04f5e0e21142b 100644
--- a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md
+++ b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README.md
@@ -49,7 +49,37 @@ tags:
-### 方法一
+### 方法一：字典树计数 + 贪心构造
+
+本题要求在区间 $[1, n]$ 中，按**字典序**排序后，找到第 $k$ 小的数字。由于 $n$ 的范围非常大（最多可达 $10^9$），我们无法直接枚举所有数字后排序。因此我们采用**贪心 + 字典树模拟**的策略。
+
+我们将 $[1, n]$ 看作一棵 **十叉字典树（Trie）**：
+
+- 每个节点是一个前缀，根节点为空串；
+- 节点的子节点是当前前缀拼接上 $0 \sim 9$；
+- 例如前缀 $1$ 会有子节点 $10, 11, \ldots, 19$，而 $10$ 会有 $100, 101, \ldots, 109$；
+- 这种结构天然符合字典序遍历。
+
+```
+根
+├── 1
+│ ├── 10
+│ ├── 11
+│ ├── ...
+├── 2
+├── ...
+```
+
+我们使用变量 $\textit{curr}$ 表示当前前缀，初始为 $1$。每次我们尝试向下扩展前缀，直到找到第 $k$ 小的数字。
+
+每次我们计算当前前缀下有多少个合法数字（即以 $\textit{curr}$ 为前缀、且不超过 $n$ 的整数个数），记作 $\textit{count}(\text{curr})$：
+
+- 如果 $k \ge \text{count}(\text{curr})$：说明目标不在这棵子树中，跳过整棵子树，前缀右移：$\textit{curr} \leftarrow \text{curr} + 1$，并更新 $k \leftarrow k - \text{count}(\text{curr})$；
+- 否则：说明目标在当前前缀的子树中，进入下一层：$\textit{curr} \leftarrow \text{curr} \times 10$，并消耗一个前缀：$k \leftarrow k - 1$。
+
+每一层我们将当前区间扩大 $10$ 倍，向下延伸到更长的前缀，直到超出 $n$。
+
+时间复杂度 $O(\log^2 n)$，空间复杂度 $O(1)$。
@@ -181,6 +211,75 @@ func findKthNumber(n int, k int) int {
}
```
+#### TypeScript
+
+```ts
+function findKthNumber(n: number, k: number): number {
+ function count(curr: number): number {
+ let next = curr + 1;
+ let cnt = 0;
+ while (curr <= n) {
+ cnt += Math.min(n - curr + 1, next - curr);
+ curr *= 10;
+ next *= 10;
+ }
+ return cnt;
+ }
+
+ let curr = 1;
+ k--;
+
+ while (k > 0) {
+ const cnt = count(curr);
+ if (k >= cnt) {
+ k -= cnt;
+ curr += 1;
+ } else {
+ k -= 1;
+ curr *= 10;
+ }
+ }
+
+ return curr;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+ pub fn find_kth_number(n: i32, k: i32) -> i32 {
+ fn count(mut curr: i64, n: i32) -> i32 {
+ let mut next = curr + 1;
+ let mut total = 0;
+ let n = n as i64;
+ while curr <= n {
+ total += std::cmp::min(n - curr + 1, next - curr);
+ curr *= 10;
+ next *= 10;
+ }
+ total as i32
+ }
+
+ let mut curr = 1;
+ let mut k = k - 1;
+
+ while k > 0 {
+ let cnt = count(curr as i64, n);
+ if k >= cnt {
+ k -= cnt;
+ curr += 1;
+ } else {
+ k -= 1;
+ curr *= 10;
+ }
+ }
+
+ curr
+ }
+}
+```
+
diff --git a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md
index f8c7bc38a68b3..abdae7d0cad00 100644
--- a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md
+++ b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/README_EN.md
@@ -47,7 +47,46 @@ tags:
-### Solution 1
+### Solution 1: Trie-Based Counting + Greedy Construction
+
+The problem asks for the \$k\$-th smallest number in the range $[1, n]$ when all numbers are sorted in **lexicographical order**. Since $n$ can be as large as $10^9$, we cannot afford to generate and sort all the numbers explicitly. Instead, we adopt a strategy based on **greedy traversal over a conceptual Trie**.
+
+We treat the range $[1, n]$ as a **10-ary prefix tree (Trie)**:
+
+- Each node represents a numeric prefix, starting from an empty root;
+- Each node has 10 children, corresponding to appending digits $0 \sim 9$;
+- For example, prefix $1$ has children $10, 11, \ldots, 19$, and node $10$ has children $100, 101, \ldots, 109$;
+- This tree naturally reflects lexicographical order traversal.
+
+```
+root
+├── 1
+│ ├── 10
+│ ├── 11
+│ ├── ...
+├── 2
+├── ...
+```
+
+We use a variable $\textit{curr}$ to denote the current prefix, initialized as $1$. At each step, we try to expand or skip prefixes until we find the \$k\$-th smallest number.
+
+At each step, we calculate how many valid numbers (i.e., numbers $\le n$ with prefix $\textit{curr}$) exist under this prefix subtree. Let this count be $\textit{count}(\text{curr})$:
+
+- If $k \ge \text{count}(\text{curr})$: the target number is not in this subtree. We skip the entire subtree by moving to the next sibling:
+
+ $$
+ \textit{curr} \leftarrow \textit{curr} + 1,\quad k \leftarrow k - \text{count}(\text{curr})
+ $$
+
+- Otherwise: the target is within this subtree. We go one level deeper:
+
+ $$
+ \textit{curr} \leftarrow \textit{curr} \times 10,\quad k \leftarrow k - 1
+ $$
+
+At each level, we enlarge the current range by multiplying by 10 and continue descending until we exceed $n$.
+
+The time complexity is $O(\log^2 n)$, as we perform logarithmic operations for counting and traversing the Trie structure. The space complexity is $O(1)$ since we only use a few variables to track the current prefix and count.
@@ -179,6 +218,75 @@ func findKthNumber(n int, k int) int {
}
```
+#### TypeScript
+
+```ts
+function findKthNumber(n: number, k: number): number {
+ function count(curr: number): number {
+ let next = curr + 1;
+ let cnt = 0;
+ while (curr <= n) {
+ cnt += Math.min(n - curr + 1, next - curr);
+ curr *= 10;
+ next *= 10;
+ }
+ return cnt;
+ }
+
+ let curr = 1;
+ k--;
+
+ while (k > 0) {
+ const cnt = count(curr);
+ if (k >= cnt) {
+ k -= cnt;
+ curr += 1;
+ } else {
+ k -= 1;
+ curr *= 10;
+ }
+ }
+
+ return curr;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+ pub fn find_kth_number(n: i32, k: i32) -> i32 {
+ fn count(mut curr: i64, n: i32) -> i32 {
+ let mut next = curr + 1;
+ let mut total = 0;
+ let n = n as i64;
+ while curr <= n {
+ total += std::cmp::min(n - curr + 1, next - curr);
+ curr *= 10;
+ next *= 10;
+ }
+ total as i32
+ }
+
+ let mut curr = 1;
+ let mut k = k - 1;
+
+ while k > 0 {
+ let cnt = count(curr as i64, n);
+ if k >= cnt {
+ k -= cnt;
+ curr += 1;
+ } else {
+ k -= 1;
+ curr *= 10;
+ }
+ }
+
+ curr
+ }
+}
+```
+
diff --git a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.rs b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.rs
new file mode 100644
index 0000000000000..52351a5f8ff07
--- /dev/null
+++ b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.rs
@@ -0,0 +1,31 @@
+impl Solution {
+ pub fn find_kth_number(n: i32, k: i32) -> i32 {
+ fn count(mut curr: i64, n: i32) -> i32 {
+ let mut next = curr + 1;
+ let mut total = 0;
+ let n = n as i64;
+ while curr <= n {
+ total += std::cmp::min(n - curr + 1, next - curr);
+ curr *= 10;
+ next *= 10;
+ }
+ total as i32
+ }
+
+ let mut curr = 1;
+ let mut k = k - 1;
+
+ while k > 0 {
+ let cnt = count(curr as i64, n);
+ if k >= cnt {
+ k -= cnt;
+ curr += 1;
+ } else {
+ k -= 1;
+ curr *= 10;
+ }
+ }
+
+ curr
+ }
+}
diff --git a/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.ts b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.ts
new file mode 100644
index 0000000000000..7e54456e1b47a
--- /dev/null
+++ b/solution/0400-0499/0440.K-th Smallest in Lexicographical Order/Solution.ts
@@ -0,0 +1,28 @@
+function findKthNumber(n: number, k: number): number {
+ function count(curr: number): number {
+ let next = curr + 1;
+ let cnt = 0;
+ while (curr <= n) {
+ cnt += Math.min(n - curr + 1, next - curr);
+ curr *= 10;
+ next *= 10;
+ }
+ return cnt;
+ }
+
+ let curr = 1;
+ k--;
+
+ while (k > 0) {
+ const cnt = count(curr);
+ if (k >= cnt) {
+ k -= cnt;
+ curr += 1;
+ } else {
+ k -= 1;
+ curr *= 10;
+ }
+ }
+
+ return curr;
+}
diff --git a/solution/0500-0599/0581.Shortest Unsorted Continuous Subarray/README.md b/solution/0500-0599/0581.Shortest Unsorted Continuous Subarray/README.md
index 6d3765e61f6cb..2cf6f98195019 100644
--- a/solution/0500-0599/0581.Shortest Unsorted Continuous Subarray/README.md
+++ b/solution/0500-0599/0581.Shortest Unsorted Continuous Subarray/README.md
@@ -173,32 +173,20 @@ function findUnsortedSubarray(nums: number[]): number {
```rust
impl Solution {
pub fn find_unsorted_subarray(nums: Vec假设 Andy 和 Doris 想在晚餐时选择一家餐厅，并且他们都有一个表示最喜爱餐厅的列表，每个餐厅的名字用字符串表示。
+给定两个字符串数组 list1 和 list2，找到 索引和最小的公共字符串。
你需要帮助他们用最少的索引和找出他们共同喜爱的餐厅。 如果答案不止一个，则输出所有答案并且不考虑顺序。 你可以假设答案总是存在。
+公共字符串 是同时出现在 list1 和 list2 中的字符串。
具有 最小索引和的公共字符串 是指，如果它在 list1[i] 和 list2[j] 中出现，那么 i + j 应该是所有其他 公共字符串 中的最小值。
返回所有 具有最小索引和的公共字符串。以 任何顺序 返回答案。
@@ -29,7 +33,7 @@ tags:
输入: list1 = ["Shogun", "Tapioca Express", "Burger King", "KFC"]，list2 = ["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"] 输出: ["Shogun"] -解释: 他们唯一共同喜爱的餐厅是“Shogun”。 +解释: 唯一的公共字符串是 “Shogun”。
示例 2:
@@ -37,9 +41,20 @@ tags:输入:list1 = ["Shogun", "Tapioca Express", "Burger King", "KFC"]，list2 = ["KFC", "Shogun", "Burger King"] 输出: ["Shogun"] -解释: 他们共同喜爱且具有最小索引和的餐厅是“Shogun”，它有最小的索引和1(0+1)。 +解释: 具有最小索引和的公共字符串是 “Shogun”，它有最小的索引和 = (0 + 1) = 1。+
示例 3：
+ ++输入：list1 = ["happy","sad","good"], list2 = ["sad","happy","good"] +输出：["sad","happy"] +解释：有三个公共字符串： +"happy" 索引和 = (0 + 1) = 1. +"sad" 索引和 = (1 + 0) = 1. +"good" 索引和 = (2 + 2) = 4. +最小索引和的字符串是 "sad" 和 "happy"。+
提示:
diff --git a/solution/0600-0699/0636.Exclusive Time of Functions/README.md b/solution/0600-0699/0636.Exclusive Time of Functions/README.md index 9dc839b951335..7f5e3268f6a69 100644 --- a/solution/0600-0699/0636.Exclusive Time of Functions/README.md +++ b/solution/0600-0699/0636.Exclusive Time of Functions/README.md @@ -73,7 +73,7 @@ tags:1 <= n <= 1001 <= logs.length <= 5002 <= logs.length <= 5000 <= function_id < n0 <= timestamp <= 1091 <= n <= 1001 <= logs.length <= 5002 <= logs.length <= 5000 <= function_id < n0 <= timestamp <= 109Input: n = 3 Output: 5 -Explanation: The five different ways are show above. +Explanation: The five different ways are shown above.
Example 2:
@@ -126,12 +126,11 @@ class Solution { ```cpp class Solution { public: - const int mod = 1e9 + 7; - int numTilings(int n) { - long f[4] = {1, 0, 0, 0}; + const int mod = 1e9 + 7; + long long f[4] = {1, 0, 0, 0}; for (int i = 1; i <= n; ++i) { - long g[4] = {0, 0, 0, 0}; + long long g[4]; g[0] = (f[0] + f[1] + f[2] + f[3]) % mod; g[1] = (f[2] + f[3]) % mod; g[2] = (f[1] + f[3]) % mod; @@ -162,6 +161,46 @@ func numTilings(n int) int { } ``` +#### TypeScript + +```ts +function numTilings(n: number): number { + const mod = 1_000_000_007; + let f: number[] = [1, 0, 0, 0]; + + for (let i = 1; i <= n; ++i) { + const g: number[] = Array(4); + g[0] = (f[0] + f[1] + f[2] + f[3]) % mod; + g[1] = (f[2] + f[3]) % mod; + g[2] = (f[1] + f[3]) % mod; + g[3] = f[0] % mod; + f = g; + } + + return f[0]; +} +``` + +#### Rust + +```rust +impl Solution { + pub fn num_tilings(n: i32) -> i32 { + const MOD: i64 = 1_000_000_007; + let mut f: [i64; 4] = [1, 0, 0, 0]; + for _ in 1..=n { + let mut g = [0i64; 4]; + g[0] = (f[0] + f[1] + f[2] + f[3]) % MOD; + g[1] = (f[2] + f[3]) % MOD; + g[2] = (f[1] + f[3]) % MOD; + g[3] = f[0] % MOD; + f = g; + } + f[0] as i32 + } +} +``` + diff --git a/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.cpp b/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.cpp index 7dee343d57916..835d82ee5f184 100644 --- a/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.cpp +++ b/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.cpp @@ -1,11 +1,10 @@ class Solution { public: - const int mod = 1e9 + 7; - int numTilings(int n) { - long f[4] = {1, 0, 0, 0}; + const int mod = 1e9 + 7; + long long f[4] = {1, 0, 0, 0}; for (int i = 1; i <= n; ++i) { - long g[4] = {0, 0, 0, 0}; + long long g[4]; g[0] = (f[0] + f[1] + f[2] + f[3]) % mod; g[1] = (f[2] + f[3]) % mod; g[2] = (f[1] + f[3]) % mod; diff --git a/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.rs b/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.rs new file mode 100644 index 0000000000000..fa193c65eb056 --- /dev/null +++ b/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.rs @@ -0,0 +1,15 @@ +impl Solution { + pub fn num_tilings(n: i32) -> i32 { + const MOD: i64 = 1_000_000_007; + let mut f: [i64; 4] = [1, 0, 0, 0]; + for _ in 1..=n { + let mut g = [0i64; 4]; + g[0] = (f[0] + f[1] + f[2] + f[3]) % MOD; + g[1] = (f[2] + f[3]) % MOD; + g[2] = (f[1] + f[3]) % MOD; + g[3] = f[0] % MOD; + f = g; + } + f[0] as i32 + } +} diff --git a/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.ts b/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.ts new file mode 100644 index 0000000000000..1550567f5375c --- /dev/null +++ b/solution/0700-0799/0790.Domino and Tromino Tiling/Solution.ts @@ -0,0 +1,15 @@ +function numTilings(n: number): number { + const mod = 1_000_000_007; + let f: number[] = [1, 0, 0, 0]; + + for (let i = 1; i <= n; ++i) { + const g: number[] = Array(4); + g[0] = (f[0] + f[1] + f[2] + f[3]) % mod; + g[1] = (f[2] + f[3]) % mod; + g[2] = (f[1] + f[3]) % mod; + g[3] = f[0] % mod; + f = g; + } + + return f[0]; +} diff --git a/solution/0800-0899/0802.Find Eventual Safe States/README.md b/solution/0800-0899/0802.Find Eventual Safe States/README.md index efac1c82f3f3f..06994079ec0b8 100644 --- a/solution/0800-0899/0802.Find Eventual Safe States/README.md +++ b/solution/0800-0899/0802.Find Eventual Safe States/README.md @@ -21,7 +21,7 @@ tags:有一个有 n 个节点的有向图，节点按 0 到 n - 1 编号。图由一个 索引从 0 开始 的 2D 整数数组 graph表示， graph[i]是与节点 i 相邻的节点的整数数组，这意味着从节点 i 到 graph[i]中的每个节点都有一条边。
如果一个节点没有连出的有向边，则该节点是 终端节点 。如果从该节点开始的所有可能路径都通向 终端节点 ，则该节点为 安全节点 。
+如果一个节点没有连出的有向边，则该节点是 终端节点 。如果从该节点开始的所有可能路径都通向 终端节点 ，则该节点为 终端节点（或另一个安全节点）。
返回一个由图中所有 安全节点 组成的数组作为答案。答案数组中的元素应当按 升序 排列。
diff --git a/solution/0800-0899/0819.Most Common Word/README.md b/solution/0800-0899/0819.Most Common Word/README.md index f3adddcb4620e..20a453ae26ab5 100644 --- a/solution/0800-0899/0819.Most Common Word/README.md +++ b/solution/0800-0899/0819.Most Common Word/README.md @@ -23,6 +23,8 @@ tags:paragraph 中的单词 不区分大小写 ，答案应以 小写 形式返回。
注意 单词不包含标点符号。
+
示例 1：
diff --git a/solution/0800-0899/0819.Most Common Word/README_EN.md b/solution/0800-0899/0819.Most Common Word/README_EN.md index ebcacda3c5c8e..2aae27ab03f78 100644 --- a/solution/0800-0899/0819.Most Common Word/README_EN.md +++ b/solution/0800-0899/0819.Most Common Word/README_EN.md @@ -23,6 +23,8 @@ tags:The words in paragraph are case-insensitive and the answer should be returned in lowercase.
Note that words can not contain punctuation symbols.
+
Example 1:
diff --git a/solution/0800-0899/0838.Push Dominoes/README.md b/solution/0800-0899/0838.Push Dominoes/README.md index e6f79c7abba0b..d37d4feea673e 100644 --- a/solution/0800-0899/0838.Push Dominoes/README.md +++ b/solution/0800-0899/0838.Push Dominoes/README.md @@ -68,7 +68,30 @@ tags: -### 方法一 +### 方法一：多源 BFS + +把所有初始受到推力的骨牌（`L` 或 `R`）视作 **源点**，它们会同时向外扩散各自的力。用队列按时间层级（0, 1, 2 …）进行 BFS： + +我们定义 $\text{time[i]}$ 记录第 *i* 张骨牌第一次受力的时刻，`-1` 表示尚未受力，定义 $\text{force[i]}$ 是一个长度可变的列表，存放该骨牌在同一时刻收到的方向（`'L'`、`'R'`）。初始时把所有 `L/R` 的下标压入队列，并将它们的时间置 0。 + +当弹出下标 *i* 时，若 $\text{force[i]}$ 只有一个方向，骨牌就会倒向该方向 $f$。设下一张骨牌下标为 + +$$ +j = +\begin{cases} +i - 1, & f = L,\\ +i + 1, & f = R. +\end{cases} +$$ + +若 $0 \leq j < n$： + +- 若 $\text{time[j]}=-1$，说明 *j* 从未受力，记录 $\text{time[j]}=\text{time[i]}+1$ 并入队，同时把 $f$ 写入 $\text{force[j]}$。 +- 若 $\text{time[j]}=\text{time[i]}+1$，说明它在同一“下一刻”已受过另一股力，此时只把 $f$ 追加到 $\text{force[j]}$，形成对冲；后续因 `len(force[j])==2`，它将保持竖直。 + +队列清空后，所有 $\text{force[i]}$ 长度为 1 的位置倒向对应方向；长度为 2 的位置保持 `.`。最终将字符数组拼接为答案。 + +时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是骨牌的数量。 @@ -242,44 +265,40 @@ func pushDominoes(dominoes string) string { ```ts function pushDominoes(dominoes: string): string { const n = dominoes.length; - const map = { - L: -1, - R: 1, - '.': 0, - }; - let ans = new Array(n).fill(0); - let visited = new Array(n).fill(0); - let queue = []; - let depth = 1; + const q: number[] = []; + const time: number[] = Array(n).fill(-1); + const force: string[][] = Array.from({ length: n }, () => []); + for (let i = 0; i < n; i++) { - let cur = map[dominoes.charAt(i)]; - if (cur) { - queue.push(i); - visited[i] = depth; - ans[i] = cur; + const f = dominoes[i]; + if (f !== '.') { + q.push(i); + time[i] = 0; + force[i].push(f); } } - while (queue.length) { - depth++; - let nextLevel = []; - for (let i of queue) { - const dx = ans[i]; - let x = i + dx; - if (x >= 0 && x < n && [0, depth].includes(visited[x])) { - ans[x] += dx; - visited[x] = depth; - nextLevel.push(x); + + const ans: string[] = Array(n).fill('.'); + let head = 0; + while (head < q.length) { + const i = q[head++]; + if (force[i].length === 1) { + const f = force[i][0]; + ans[i] = f; + const j = f === 'L' ? i - 1 : i + 1; + if (j >= 0 && j < n) { + const t = time[i]; + if (time[j] === -1) { + q.push(j); + time[j] = t + 1; + force[j].push(f); + } else if (time[j] === t + 1) { + force[j].push(f); + } } } - queue = nextLevel; } - return ans - .map(d => { - if (!d) return '.'; - else if (d < 0) return 'L'; - else return 'R'; - }) - .join(''); + return ans.join(''); } ``` diff --git a/solution/0800-0899/0838.Push Dominoes/README_EN.md b/solution/0800-0899/0838.Push Dominoes/README_EN.md index ce232677d18a7..ec5ef4c80d36c 100644 --- a/solution/0800-0899/0838.Push Dominoes/README_EN.md +++ b/solution/0800-0899/0838.Push Dominoes/README_EN.md @@ -67,7 +67,30 @@ tags: -### Solution 1 +### Solution 1: Multi-Source BFS + +Treat all initially pushed dominoes (`L` or `R`) as **sources**, which simultaneously propagate their forces outward. Use a queue to perform BFS layer by layer (0, 1, 2, ...): + +We define $\text{time[i]}$ to record the first moment when the _i_-th domino is affected by a force, with `-1` indicating it has not been affected yet. We also define $\text{force[i]}$ as a variable-length list that stores the directions (`'L'`, `'R'`) of forces acting on the domino at the same moment. Initially, push all indices of `L/R` dominoes into the queue and set their `time` to 0. + +When dequeuing index _i_, if $\text{force[i]}$ contains only one direction, the domino will fall in that direction $f$. Let the index of the next domino be: + +$$ +j = +\begin{cases} +i - 1, & f = L,\\ +i + 1, & f = R. +\end{cases} +$$ + +If $0 \leq j < n$: + +- If $\text{time[j]} = -1$, it means _j_ has not been affected yet. Record $\text{time[j]} = \text{time[i]} + 1$, enqueue it, and append $f$ to $\text{force[j]}$. +- If $\text{time[j]} = \text{time[i]} + 1$, it means _j_ has already been affected by another force at the same "next moment." In this case, append $f$ to $\text{force[j]}$, causing a standoff. Subsequently, since $\text{len(force[j])} = 2$, it will remain upright. + +After the queue is emptied, all positions where $\text{force[i]}$ has a length of 1 will fall in the corresponding direction, while positions with a length of 2 will remain as `.`. Finally, concatenate the character array to form the answer. + +The complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of dominoes. @@ -241,44 +264,40 @@ func pushDominoes(dominoes string) string { ```ts function pushDominoes(dominoes: string): string { const n = dominoes.length; - const map = { - L: -1, - R: 1, - '.': 0, - }; - let ans = new Array(n).fill(0); - let visited = new Array(n).fill(0); - let queue = []; - let depth = 1; + const q: number[] = []; + const time: number[] = Array(n).fill(-1); + const force: string[][] = Array.from({ length: n }, () => []); + for (let i = 0; i < n; i++) { - let cur = map[dominoes.charAt(i)]; - if (cur) { - queue.push(i); - visited[i] = depth; - ans[i] = cur; + const f = dominoes[i]; + if (f !== '.') { + q.push(i); + time[i] = 0; + force[i].push(f); } } - while (queue.length) { - depth++; - let nextLevel = []; - for (let i of queue) { - const dx = ans[i]; - let x = i + dx; - if (x >= 0 && x < n && [0, depth].includes(visited[x])) { - ans[x] += dx; - visited[x] = depth; - nextLevel.push(x); + + const ans: string[] = Array(n).fill('.'); + let head = 0; + while (head < q.length) { + const i = q[head++]; + if (force[i].length === 1) { + const f = force[i][0]; + ans[i] = f; + const j = f === 'L' ? i - 1 : i + 1; + if (j >= 0 && j < n) { + const t = time[i]; + if (time[j] === -1) { + q.push(j); + time[j] = t + 1; + force[j].push(f); + } else if (time[j] === t + 1) { + force[j].push(f); + } } } - queue = nextLevel; } - return ans - .map(d => { - if (!d) return '.'; - else if (d < 0) return 'L'; - else return 'R'; - }) - .join(''); + return ans.join(''); } ``` diff --git a/solution/0800-0899/0838.Push Dominoes/Solution.ts b/solution/0800-0899/0838.Push Dominoes/Solution.ts index d9e8412c5d062..0b912d31ca203 100644 --- a/solution/0800-0899/0838.Push Dominoes/Solution.ts +++ b/solution/0800-0899/0838.Push Dominoes/Solution.ts @@ -1,41 +1,37 @@ function pushDominoes(dominoes: string): string { const n = dominoes.length; - const map = { - L: -1, - R: 1, - '.': 0, - }; - let ans = new Array(n).fill(0); - let visited = new Array(n).fill(0); - let queue = []; - let depth = 1; + const q: number[] = []; + const time: number[] = Array(n).fill(-1); + const force: string[][] = Array.from({ length: n }, () => []); + for (let i = 0; i < n; i++) { - let cur = map[dominoes.charAt(i)]; - if (cur) { - queue.push(i); - visited[i] = depth; - ans[i] = cur; + const f = dominoes[i]; + if (f !== '.') { + q.push(i); + time[i] = 0; + force[i].push(f); } } - while (queue.length) { - depth++; - let nextLevel = []; - for (let i of queue) { - const dx = ans[i]; - let x = i + dx; - if (x >= 0 && x < n && [0, depth].includes(visited[x])) { - ans[x] += dx; - visited[x] = depth; - nextLevel.push(x); + + const ans: string[] = Array(n).fill('.'); + let head = 0; + while (head < q.length) { + const i = q[head++]; + if (force[i].length === 1) { + const f = force[i][0]; + ans[i] = f; + const j = f === 'L' ? i - 1 : i + 1; + if (j >= 0 && j < n) { + const t = time[i]; + if (time[j] === -1) { + q.push(j); + time[j] = t + 1; + force[j].push(f); + } else if (time[j] === t + 1) { + force[j].push(f); + } } } - queue = nextLevel; } - return ans - .map(d => { - if (!d) return '.'; - else if (d < 0) return 'L'; - else return 'R'; - }) - .join(''); + return ans.join(''); } diff --git a/solution/0900-0999/0909.Snakes and Ladders/README.md b/solution/0900-0999/0909.Snakes and Ladders/README.md index f260b58f25623..f5feac9a8def4 100644 --- a/solution/0900-0999/0909.Snakes and Ladders/README.md +++ b/solution/0900-0999/0909.Snakes and Ladders/README.md @@ -279,6 +279,52 @@ function snakesAndLadders(board: number[][]): number { } ``` +#### Rust + +```rust +use std::collections::{HashSet, VecDeque}; + +impl Solution { + pub fn snakes_and_ladders(board: Vec输入：s1 = "parker", s2 = "morris", baseStr = "parser" 输出："makkek" -解释：根据A和B 中的等价信息，我们可以将这些字符分为[m,p],[a,o],[k,r,s],[e,i] 共 4 组。每组中的字符都是等价的，并按字典序排列。所以答案是"makkek"。 +解释：根据A和B中的等价信息，我们可以将这些字符分为[m,p],[a,o],[k,r,s],[e,i]共 4 组。每组中的字符都是等价的，并按字典序排列。所以答案是"makkek"。
示例 2：
@@ -52,7 +52,7 @@ tags:输入：s1 = "hello", s2 = "world", baseStr = "hold" 输出："hdld" -解释：根据A和B 中的等价信息，我们可以将这些字符分为[h,w],[d,e,o],[l,r] 共 3 组。所以只有 S 中的第二个字符'o'变成'd'，最后答案为"hdld"。 +解释：根据A和B中的等价信息，我们可以将这些字符分为[h,w],[d,e,o],[l,r]共 3 组。所以只有 S 中的第二个字符'o'变成'd'，最后答案为"hdld"。
示例 3：
@@ -60,7 +60,7 @@ tags:输入：s1 = "leetcode", s2 = "programs", baseStr = "sourcecode" 输出："aauaaaaada" -解释：我们可以把 A 和 B 中的等价字符分为[a,o,e,r,s,c],[l,p],[g,t]和[d,m] 共 4 组，因此S中除了'u'和'd'之外的所有字母都转化成了'a'，最后答案为"aauaaaaada"。 +解释：我们可以把A和B中的等价字符分为[a,o,e,r,s,c],[l,p],[g,t]和[d,m]共 4 组，因此S中除了'u'和'd'之外的所有字母都转化成了'a'，最后答案为"aauaaaaada"。
@@ -79,7 +79,11 @@ tags: -### 方法一 +### 方法一：并查集 + +我们可以使用并查集来处理等价字符的关系。每个字符可以看作一个节点，等价关系可以看作是连接这些节点的边。通过并查集，我们可以将所有等价的字符归为一类，并且在查询时能够快速找到每个字符的代表元素。我们在进行合并操作时，始终将代表元素设置为字典序最小的字符，这样可以确保最终得到的字符串是按字典序排列的最小等价字符串。 + +时间复杂度 $O((n + m) \times \log |\Sigma|)$，空间复杂度 $O(|\Sigma|)$。其中 $n$ 是字符串 $s1$ 和 $s2$ 的长度，而 $m$ 是字符串 $baseStr$ 的长度，而 $|\Sigma|$ 是字符集的大小，本题中 $|\Sigma| = 26$。 @@ -88,54 +92,47 @@ tags: ```python class Solution: def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str: - p = list(range(26)) - - def find(x): + def find(x: int) -> int: if p[x] != x: p[x] = find(p[x]) return p[x] - for i in range(len(s1)): - a, b = ord(s1[i]) - ord('a'), ord(s2[i]) - ord('a') - pa, pb = find(a), find(b) - if pa < pb: - p[pb] = pa + p = list(range(26)) + for a, b in zip(s1, s2): + x, y = ord(a) - ord("a"), ord(b) - ord("a") + px, py = find(x), find(y) + if px < py: + p[py] = px else: - p[pa] = pb - - res = [] - for a in baseStr: - a = ord(a) - ord('a') - res.append(chr(find(a) + ord('a'))) - return ''.join(res) + p[px] = py + return "".join(chr(find(ord(c) - ord("a")) + ord("a")) for c in baseStr) ``` #### Java ```java class Solution { - private int[] p; + private final int[] p = new int[26]; public String smallestEquivalentString(String s1, String s2, String baseStr) { - p = new int[26]; - for (int i = 0; i < 26; ++i) { + for (int i = 0; i < p.length; ++i) { p[i] = i; } for (int i = 0; i < s1.length(); ++i) { - int a = s1.charAt(i) - 'a', b = s2.charAt(i) - 'a'; - int pa = find(a), pb = find(b); - if (pa < pb) { - p[pb] = pa; + int x = s1.charAt(i) - 'a'; + int y = s2.charAt(i) - 'a'; + int px = find(x), py = find(y); + if (px < py) { + p[py] = px; } else { - p[pa] = pb; + p[px] = py; } } - StringBuilder sb = new StringBuilder(); - for (char a : baseStr.toCharArray()) { - char b = (char) (find(a - 'a') + 'a'); - sb.append(b); + char[] s = baseStr.toCharArray(); + for (int i = 0; i < s.length; ++i) { + s[i] = (char) ('a' + find(s[i] - 'a')); } - return sb.toString(); + return String.valueOf(s); } private int find(int x) { @@ -152,32 +149,30 @@ class Solution { ```cpp class Solution { public: - vector
- -
产品表 Product：
-+--------------+---------+ -| Column Name | Type | -+--------------+---------+ -| product_id | int | -| product_name | varchar | -+--------------+---------+ -product_id 是这张表的主键（具有唯一值的列）。 -这张表的每一行都标识：每个产品的 id 和 产品名称。+
编写解决方案，选出每个售出过的产品 第一年 销售的 产品 id、年份、数量 和 价格。
-+
product_id，找到其在Sales表中首次出现的最早年份。编写解决方案，选出每个售出过的产品 第一年 销售的 产品 id、年份、数量 和 价格。
+返回一张有这些列的表：product_id，first_year，quantity 和 price。
结果表中的条目可以按 任意顺序 排列。
-结果格式如下例所示：
-
示例 1：
@@ -70,14 +60,6 @@ Sales 表： | 2 | 100 | 2009 | 12 | 5000 | | 7 | 200 | 2011 | 15 | 9000 | +---------+------------+------+----------+-------+ -Product 表： -+------------+--------------+ -| product_id | product_name | -+------------+--------------+ -| 100 | Nokia | -| 200 | Apple | -| 300 | Samsung | -+------------+--------------+ 输出： +------------+------------+----------+-------+ | product_id | first_year | quantity | price | diff --git a/solution/1000-1099/1070.Product Sales Analysis III/README_EN.md b/solution/1000-1099/1070.Product Sales Analysis III/README_EN.md index 03c8643cae525..c948fe652735b 100644 --- a/solution/1000-1099/1070.Product Sales Analysis III/README_EN.md +++ b/solution/1000-1099/1070.Product Sales Analysis III/README_EN.md @@ -30,32 +30,25 @@ tags: +-------------+-------+ (sale_id, year) is the primary key (combination of columns with unique values) of this table. product_id is a foreign key (reference column) toProduct table.
-Each row of this table shows a sale on the product product_id in a certain year.
-Note that the price is per unit.
-
-
-- -
Table: Product
-+--------------+---------+ -| Column Name | Type | -+--------------+---------+ -| product_id | int | -| product_name | varchar | -+--------------+---------+ -product_id is the primary key (column with unique values) of this table. -Each row of this table indicates the product name of each product.-
- -
Write a solution to select the product id, year, quantity, and price for the first year of every product sold.
+Write a solution to find all sales that occurred in the first year each product was sold.
-Return the resulting table in any order.
+For each product_id, identify the earliest year it appears in the Sales table.
Return all sales entries for that product in that year.
+The result format is in the following example.
+Return a table with the following columns: product_id, first_year, quantity, and price.
+Return the result in any order.
Example 1:
@@ -70,14 +63,7 @@ Sales table: | 2 | 100 | 2009 | 12 | 5000 | | 7 | 200 | 2011 | 15 | 9000 | +---------+------------+------+----------+-------+ -Product table: -+------------+--------------+ -| product_id | product_name | -+------------+--------------+ -| 100 | Nokia | -| 200 | Apple | -| 300 | Samsung | -+------------+--------------+ + Output: +------------+------------+----------+-------+ | product_id | first_year | quantity | price | diff --git a/solution/1100-1199/1128.Number of Equivalent Domino Pairs/README.md b/solution/1100-1199/1128.Number of Equivalent Domino Pairs/README.md index 1929c905a49f2..42812ca2724da 100644 --- a/solution/1100-1199/1128.Number of Equivalent Domino Pairs/README.md +++ b/solution/1100-1199/1128.Number of Equivalent Domino Pairs/README.md @@ -62,7 +62,7 @@ tags: 我们可以将每个多米诺骨牌的两个数字按照大小顺序拼接成一个两位数，这样就可以将等价的多米诺骨牌拼接成相同的两位数。例如，`[1, 2]` 和 `[2, 1]` 拼接成的两位数都是 `12`，`[3, 4]` 和 `[4, 3]` 拼接成的两位数都是 `34`。 -然后我们遍历所有的多米诺骨牌，用一个长度为 $100$ 的数组 $cnt$ 记录每个两位数出现的次数。对于每个多米诺骨牌，我们拼接成的两位数为 $x$，那么答案就会增加 $cnt[x]$，接着我们将 $cnt[x]$ 的值加 $1$。继续遍历下一个多米诺骨牌，就可以统计出所有等价的多米诺骨牌对的数量。 +然后我们遍历所有的多米诺骨牌，用一个哈希表或者一个长度为 $100$ 的数组 $cnt$ 记录每个两位数出现的次数。对于每个多米诺骨牌，我们拼接成的两位数为 $x$，那么答案就会增加 $cnt[x]$，接着我们将 $cnt[x]$ 的值加 $1$。继续遍历下一个多米诺骨牌，就可以统计出所有等价的多米诺骨牌对的数量。 时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 是多米诺骨牌的数量，而 $C$ 是多米诺骨牌中拼接成的两位数的最大数量，即 $100$。 @@ -132,6 +132,44 @@ func numEquivDominoPairs(dominoes [][]int) (ans int) { } ``` +#### TypeScript + +```ts +function numEquivDominoPairs(dominoes: number[][]): number { + const cnt: number[] = new Array(100).fill(0); + let ans = 0; + + for (const [a, b] of dominoes) { + const key = a < b ? a * 10 + b : b * 10 + a; + ans += cnt[key]; + cnt[key]++; + } + + return ans; +} +``` + +#### Rust + +```rust +impl Solution { + pub fn num_equiv_domino_pairs(dominoes: Vec-
请查询出所有浏览过自己文章的作者
+请查询出所有浏览过自己文章的作者。
-结果按照 id 升序排列。
结果按照作者的 id 升序排列。
查询结果的格式如下所示：
diff --git a/solution/1100-1199/1160.Find Words That Can Be Formed by Characters/README.md b/solution/1100-1199/1160.Find Words That Can Be Formed by Characters/README.md index 6b9897e53caa4..d39c633d50950 100644 --- a/solution/1100-1199/1160.Find Words That Can Be Formed by Characters/README.md +++ b/solution/1100-1199/1160.Find Words That Can Be Formed by Characters/README.md @@ -21,13 +21,11 @@ tags: -给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。
给定一个字符串数组 words 和一个字符串 chars。
假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。
如果字符串可以由 chars 中的字符组成（每个字符在 每个 words 中只能使用一次），则认为它是好的。
注意：每次拼写（指拼写词汇表中的一个单词）时，chars 中的每个字母都只能用一次。
返回词汇表 words 中你掌握的所有单词的 长度之和。
返回 words 中所有好的字符串的长度之和。
@@ -56,7 +54,7 @@ tags:
1 <= words.length <= 10001 <= words[i].length, chars.length <= 100words[i] 和 chars 中都仅包含小写英文字母You are given an array of strings words and a string chars.
A string is good if it can be formed by characters from chars (each character can only be used once).
A string is good if it can be formed by characters from chars (each character can only be used once for each word in words).
Return the sum of lengths of all good strings in words.
diff --git a/solution/1200-1299/1257.Smallest Common Region/README.md b/solution/1200-1299/1257.Smallest Common Region/README.md index f9a73717dfc2d..9f9d29a65c437 100644 --- a/solution/1200-1299/1257.Smallest Common Region/README.md +++ b/solution/1200-1299/1257.Smallest Common Region/README.md @@ -23,14 +23,14 @@ tags: -给你一些区域列表 regions ，每个列表的第一个区域都包含这个列表内所有其他区域。
给你一些区域列表 regions ，每个列表的第一个区域都 直接 包含这个列表内所有其他区域。
如果一个区域 x 直接包含区域 y，并且区域 y 直接包含区域 z，那么说区域 x 间接 包含区域 z。请注意，区域 x 也 间接 包含所有在 y 中 间接 包含的区域。
很自然地，如果区域 x 包含区域 y ，那么区域 x 比区域 y 大。同时根据定义，区域 x 包含自身。
给定两个区域 region1 和 region2 ，找到同时包含这两个区域的 最小 区域。
如果给定区域 r1，r2 和 r3，使得 r1 包含 r3，那么数据保证 r2 不会包含 r3 。
数据同样保证最小区域一定存在。
diff --git a/solution/1200-1299/1257.Smallest Common Region/README_EN.md b/solution/1200-1299/1257.Smallest Common Region/README_EN.md index d5b52f58e36c7..2085a7c0a74a4 100644 --- a/solution/1200-1299/1257.Smallest Common Region/README_EN.md +++ b/solution/1200-1299/1257.Smallest Common Region/README_EN.md @@ -23,13 +23,13 @@ tags: -
You are given some lists of regions where the first region of each list includes all other regions in that list.
You are given some lists of regions where the first region of each list directly contains all other regions in that list.
Naturally, if a region x contains another region y then x is bigger than y. Also, by definition, a region x contains itself.
If a region x contains a region y directly, and region y contains region z directly, then region x is said to contain region z indirectly. Note that region x also indirectly contains all regions indirectly containd in y.
Given two regions: region1 and region2, return the smallest region that contains both of them.
Naturally, if a region x contains (either directly or indirectly) another region y, then x is bigger than or equal to y in size. Also, by definition, a region x contains itself.
If you are given regions r1, r2, and r3 such that r1 includes r3, it is guaranteed there is no r2 such that r2 includes r3.
Given two regions: region1 and region2, return the smallest region that contains both of them.
It is guaranteed the smallest region exists.
@@ -65,6 +65,7 @@ region2 = "New York"region1 != region2regions[i][j], region1, and region2 consist of English letters.给你一个整数数组 nums，请你返回其中位数为 偶数 的数字的个数。
给你一个整数数组 nums，请你返回其中包含 偶数 个数位的数字的个数。
@@ -131,6 +131,16 @@ function findNumbers(nums: number[]): number { } ``` +#### Rust + +```rust +impl Solution { + pub fn find_numbers(nums: Vec
containedBoxes[i]：整数，表示放在 box[i] 里的盒子所对应的下标。给你一个 initialBoxes 数组，表示你现在得到的盒子，你可以获得里面的糖果，也可以用盒子里的钥匙打开新的盒子，还可以继续探索从这个盒子里找到的其他盒子。
给你一个整数数组 initialBoxes，包含你最初拥有的盒子。你可以拿走每个 已打开盒子 里的所有糖果，并且可以使用其中的钥匙去开启新的盒子，并且可以使用在其中发现的其他盒子。
请你按照上述规则，返回可以获得糖果的 最大数目 。
@@ -37,7 +37,8 @@ tags:示例 1：
-
