Add solution 189

halfrost · halfrost · commit 1bc9d342090e · 2021-01-08T02:44:27.000+08:00
diff --git a/README.md b/README.md
@@ -120,6 +120,8 @@
 
 ## 一. 目录
 
+以下已经收录了 556 道题的题解，还有 13 道题在尝试优化到 beats 100%
+
 | No.    |  Title  |  Solution  |  Acceptance |  Difficulty |  Frequency |
 |:--------:|:--------------------------------------------------------------|:--------:|:--------:|:--------:|:--------:|
 |0001|Two Sum|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0001.Two-Sum)|46.1%|Easy||
@@ -310,7 +312,7 @@
 |0186|Reverse Words in a String II||45.0%|Medium||
 |0187|Repeated DNA Sequences|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0187.Repeated-DNA-Sequences)|41.2%|Medium||
 |0188|Best Time to Buy and Sell Stock IV||29.2%|Hard||
-|0189|Rotate Array||36.3%|Medium||
+|0189|Rotate Array|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0189.Rotate-Array)|36.3%|Medium||
 |0190|Reverse Bits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0190.Reverse-Bits)|41.5%|Easy||
 |0191|Number of 1 Bits|[Go](https://github.com/halfrost/LeetCode-Go/tree/master/leetcode/0191.Number-of-1-Bits)|51.8%|Easy||
 |0192|Word Frequency||25.8%|Medium||
diff --git a/automation/render.go b/automation/render.go
@@ -17,6 +17,8 @@ import (
 	"strings"
 )
 
+var try int
+
 func main() {
 	var (
 		result []m.StatStatusPairs
@@ -40,7 +42,7 @@ func main() {
 	// res, _ := json.Marshal(mdrows)
 	//writeFile("leetcode_problem", res)
 	mds := m.Mdrows{Mdrows: mdrows}
-	res, err := readFile("./template.markdown", "{{.AvailableTable}}", mds)
+	res, err := readFile("./template.markdown", "{{.AvailableTable}}", len(solutionIds), try, mds)
 	if err != nil {
 		fmt.Println(err)
 		return
@@ -109,6 +111,7 @@ func loadSolutionsDir() []int {
 		}
 	}
 	sort.Ints(solutionIds)
+	try = len(files) - len(solutionIds)
 	fmt.Printf("读取了 %v 道题的题解，当前目录下有 %v 个文件(可能包含 .DS_Store)，有 %v 道题在尝试中\n", len(solutionIds), len(files), len(files)-len(solutionIds))
 	return solutionIds
 }
@@ -153,7 +156,7 @@ func readTMPL(path string) string {
 	return string(data)
 }
 
-func readFile(filePath, template string, mdrows m.Mdrows) ([]byte, error) {
+func readFile(filePath, template string, total, try int, mdrows m.Mdrows) ([]byte, error) {
 	f, err := os.OpenFile(filePath, os.O_RDONLY, 0644)
 	if err != nil {
 		return nil, err
@@ -174,10 +177,14 @@ func readFile(filePath, template string, mdrows m.Mdrows) ([]byte, error) {
 			newByte := reg.ReplaceAll(line, []byte(mdrows.AvailableTable()))
 			output = append(output, newByte...)
 			output = append(output, []byte("\n")...)
+		} else if ok, _ := regexp.Match("{{.TotalNum}}", line); ok {
+			reg := regexp.MustCompile("{{.TotalNum}}")
+			newByte := reg.ReplaceAll(line, []byte(fmt.Sprintf("以下已经收录了 %v 道题的题解，还有 %v 道题在尝试优化到 beats 100%%", total, try)))
+			output = append(output, newByte...)
+			output = append(output, []byte("\n")...)
 		} else {
 			output = append(output, line...)
 			output = append(output, []byte("\n")...)
 		}
 	}
-	return output, nil
 }
diff --git a/automation/template.markdown b/automation/template.markdown
@@ -120,6 +120,8 @@
 
 ## 一. 目录
 
+{{.TotalNum}}
+
 {{.AvailableTable}}
 
 ------------------------------------------------------------------
diff --git a/leetcode/0189.Rotate-Array/189. Rotate Array.go b/leetcode/0189.Rotate-Array/189. Rotate Array.go
@@ -0,0 +1,24 @@
+package leetcode
+
+// 解法一 时间复杂度 O(n)，空间复杂度 O(1)
+func rotate(nums []int, k int) {
+	k %= len(nums)
+	reverse(nums)
+	reverse(nums[:k])
+	reverse(nums[k:])
+}
+
+func reverse(a []int) {
+	for i, n := 0, len(a); i < n/2; i++ {
+		a[i], a[n-1-i] = a[n-1-i], a[i]
+	}
+}
+
+// 解法二 时间复杂度 O(n)，空间复杂度 O(n)
+func rotate1(nums []int, k int) {
+	newNums := make([]int, len(nums))
+	for i, v := range nums {
+		newNums[(i+k)%len(nums)] = v
+	}
+	copy(nums, newNums)
+}
diff --git a/leetcode/0189.Rotate-Array/189. Rotate Array_test.go b/leetcode/0189.Rotate-Array/189. Rotate Array_test.go
@@ -0,0 +1,50 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question189 struct {
+	para189
+	ans189
+}
+
+// para 是参数
+// one 代表第一个参数
+type para189 struct {
+	nums []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans189 struct {
+	one []int
+}
+
+func Test_Problem189(t *testing.T) {
+
+	qs := []question189{
+
+		{
+			para189{[]int{1, 2, 3, 4, 5, 6, 7}, 3},
+			ans189{[]int{5, 6, 7, 1, 2, 3, 4}},
+		},
+
+		{
+			para189{[]int{-1, -100, 3, 99}, 2},
+			ans189{[]int{3, 99, -1, -100}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 189------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans189, q.para189
+		fmt.Printf("【input】:%v       ", p)
+		rotate(p.nums, p.k)
+		fmt.Printf("【output】:%v\n", p.nums)
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0189.Rotate-Array/README.md b/leetcode/0189.Rotate-Array/README.md
@@ -0,0 +1,75 @@
+# [189. Rotate Array](https://leetcode.com/problems/rotate-array/)
+
+## 题目
+
+Given an array, rotate the array to the right by *k* steps, where *k* is non-negative.
+
+**Follow up:**
+
+- Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
+- Could you do it in-place with O(1) extra space?
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4,5,6,7], k = 3
+Output: [5,6,7,1,2,3,4]
+Explanation:
+rotate 1 steps to the right: [7,1,2,3,4,5,6]
+rotate 2 steps to the right: [6,7,1,2,3,4,5]
+rotate 3 steps to the right: [5,6,7,1,2,3,4]
+```
+
+**Example 2:**
+
+```
+Input: nums = [-1,-100,3,99], k = 2
+Output: [3,99,-1,-100]
+Explanation: 
+rotate 1 steps to the right: [99,-1,-100,3]
+rotate 2 steps to the right: [3,99,-1,-100]
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 2 * 10^4`
+- `-2^31 <= nums[i] <= 2^31 - 1`
+- `0 <= k <= 10^5`
+
+## 题目大意
+
+给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。
+
+## 解题思路
+
+- 解法二，使用一个额外的数组，先将原数组下标为 i 的元素移动到 `(i+k) mod n` 的位置，再将剩下的元素拷贝回来即可。
+- 解法一，由于题目要求不能使用额外的空间，所以本题最佳解法不是解法二。翻转最终态是，末尾 `k mod n` 个元素移动至了数组头部，剩下的元素右移 `k mod n` 个位置至最尾部。确定了最终态以后再变换就很容易。先将数组中所有元素从头到尾翻转一次，尾部的所有元素都到了头部，然后再将 `[0,(k mod n) − 1]` 区间内的元素翻转一次，最后再将 `[k mod n, n − 1]` 区间内的元素翻转一次，即可满足题目要求。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 时间复杂度 O(n)，空间复杂度 O(1)
+func rotate(nums []int, k int) {
+	k %= len(nums)
+	reverse(nums)
+	reverse(nums[:k])
+	reverse(nums[k:])
+}
+
+func reverse(a []int) {
+	for i, n := 0, len(a); i < n/2; i++ {
+		a[i], a[n-1-i] = a[n-1-i], a[i]
+	}
+}
+
+// 解法二 时间复杂度 O(n)，空间复杂度 O(n)
+func rotate1(nums []int, k int) {
+	newNums := make([]int, len(nums))
+	for i, v := range nums {
+		newNums[(i+k)%len(nums)] = v
+	}
+	copy(nums, newNums)
+}
+```
diff --git a/website/content/ChapterFour/0189.Rotate-Array.md b/website/content/ChapterFour/0189.Rotate-Array.md
@@ -0,0 +1,75 @@
+# [189. Rotate Array](https://leetcode.com/problems/rotate-array/)
+
+## 题目
+
+Given an array, rotate the array to the right by *k* steps, where *k* is non-negative.
+
+**Follow up:**
+
+- Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
+- Could you do it in-place with O(1) extra space?
+
+**Example 1:**
+
+```
+Input: nums = [1,2,3,4,5,6,7], k = 3
+Output: [5,6,7,1,2,3,4]
+Explanation:
+rotate 1 steps to the right: [7,1,2,3,4,5,6]
+rotate 2 steps to the right: [6,7,1,2,3,4,5]
+rotate 3 steps to the right: [5,6,7,1,2,3,4]
+```
+
+**Example 2:**
+
+```
+Input: nums = [-1,-100,3,99], k = 2
+Output: [3,99,-1,-100]
+Explanation: 
+rotate 1 steps to the right: [99,-1,-100,3]
+rotate 2 steps to the right: [3,99,-1,-100]
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 2 * 10^4`
+- `-2^31 <= nums[i] <= 2^31 - 1`
+- `0 <= k <= 10^5`
+
+## 题目大意
+
+给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。
+
+## 解题思路
+
+- 解法二，使用一个额外的数组，先将原数组下标为 i 的元素移动到 `(i+k) mod n` 的位置，再将剩下的元素拷贝回来即可。
+- 解法一，由于题目要求不能使用额外的空间，所以本题最佳解法不是解法二。翻转最终态是，末尾 `k mod n` 个元素移动至了数组头部，剩下的元素右移 `k mod n` 个位置至最尾部。确定了最终态以后再变换就很容易。先将数组中所有元素从头到尾翻转一次，尾部的所有元素都到了头部，然后再将 `[0,(k mod n) − 1]` 区间内的元素翻转一次，最后再将 `[k mod n, n − 1]` 区间内的元素翻转一次，即可满足题目要求。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 时间复杂度 O(n)，空间复杂度 O(1)
+func rotate(nums []int, k int) {
+	k %= len(nums)
+	reverse(nums)
+	reverse(nums[:k])
+	reverse(nums[k:])
+}
+
+func reverse(a []int) {
+	for i, n := 0, len(a); i < n/2; i++ {
+		a[i], a[n-1-i] = a[n-1-i], a[i]
+	}
+}
+
+// 解法二 时间复杂度 O(n)，空间复杂度 O(n)
+func rotate1(nums []int, k int) {
+	newNums := make([]int, len(nums))
+	for i, v := range nums {
+		newNums[(i+k)%len(nums)] = v
+	}
+	copy(nums, newNums)
+}
+```
diff --git a/website/content/menu/index.md b/website/content/menu/index.md
@@ -168,6 +168,7 @@ headless: true
     - [0174.Dungeon-Game]({{< relref "/ChapterFour/0174.Dungeon-Game.md" >}})
     - [0179.Largest-Number]({{< relref "/ChapterFour/0179.Largest-Number.md" >}})
     - [0187.Repeated-DNA-Sequences]({{< relref "/ChapterFour/0187.Repeated-DNA-Sequences.md" >}})
+    - [0189.Rotate-Array]({{< relref "/ChapterFour/0189.Rotate-Array.md" >}})
     - [0190.Reverse-Bits]({{< relref "/ChapterFour/0190.Reverse-Bits.md" >}})
     - [0191.Number-of-1-Bits]({{< relref "/ChapterFour/0191.Number-of-1-Bits.md" >}})
     - [0198.House-Robber]({{< relref "/ChapterFour/0198.House-Robber.md" >}})
