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May 31, 2024
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feat: update lc problems #2988

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6 changes: 3 additions & 3 deletions lcof2/剑指 Offer II 005. 单词长度的最大乘积/README.md
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Expand Up @@ -188,13 +188,13 @@ class Solution {
func maxProduct(_ words: [String]) -> Int {
let n = words.count
var masks = [Int](repeating: 0, count: n)

for i in 0..<n {
for c in words[i] {
masks[i] |= 1 << (c.asciiValue! - Character("a").asciiValue!)
}
}

var maxProduct = 0
for i in 0..<n {
for j in i+1..<n {
Expand All @@ -203,7 +203,7 @@ class Solution {
}
}
}

return maxProduct
}
}
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4 changes: 2 additions & 2 deletions lcof2/剑指 Offer II 008. 和大于等于 target 的最短子数组/README.md
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Expand Up @@ -185,7 +185,7 @@ class Solution {
var ans = inf
var sum = 0
var i = 0

for j in 0..<nums.count {
sum += nums[j]
while sum >= target {
Expand All @@ -194,7 +194,7 @@ class Solution {
i += 1
}
}

return ans == inf ? 0 : ans
}
}
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6 changes: 3 additions & 3 deletions lcof2/剑指 Offer II 009. 乘积小于 K 的子数组/README.md
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Expand Up @@ -155,11 +155,11 @@ function numSubarrayProductLessThanK(nums: number[], k: number): number {
class Solution {
func numSubarrayProductLessThanK(_ nums: [Int], _ k: Int) -> Int {
if k <= 1 { return 0 }

var product: Int = 1
var ans: Int = 0
var left: Int = 0

for right in 0..<nums.count {
product *= nums[right]
while product >= k {
Expand All @@ -168,7 +168,7 @@ class Solution {
}
ans += right - left + 1
}

return ans
}
}
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4 changes: 2 additions & 2 deletions lcof2/剑指 Offer II 010. 和为 k 的子数组/README.md
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Expand Up @@ -151,13 +151,13 @@ class Solution {
var cnt: [Int: Int] = [0: 1]
var ans = 0
var s = 0

for x in nums {
s += x
ans += cnt[s - k, default: 0]
cnt[s, default: 0] += 1
}

return ans
}
}
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4 changes: 2 additions & 2 deletions lcof2/剑指 Offer II 011. 0 和 1 个数相同的子数组/README.md
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Expand Up @@ -169,7 +169,7 @@ class Solution {
var d: [Int: Int] = [0: -1]
var ans = 0
var s = 0

for i in 0..<nums.count {
s += nums[i] == 0 ? -1 : 1
if let prevIndex = d[s] {
Expand All @@ -178,7 +178,7 @@ class Solution {
d[s] = i
}
}

return ans
}
}
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2 changes: 1 addition & 1 deletion lcof2/剑指 Offer II 012. 左右两边子数组的和相等/README.md
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Expand Up @@ -233,7 +233,7 @@ class Solution {
func pivotIndex(_ nums: [Int]) -> Int {
var leftSum = 0
var rightSum = nums.reduce(0, +)

for i in 0..<nums.count {
rightSum -= nums[i]
if leftSum == rightSum {
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2 changes: 1 addition & 1 deletion lcof2/剑指 Offer II 013. 二维子矩阵的和/README.md
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Expand Up @@ -264,7 +264,7 @@ class NumMatrix {
let m = matrix.count
let n = matrix[0].count
prefixSum = Array(repeating: Array(repeating: 0, count: n + 1), count: m + 1)

for i in 1...m {
for j in 1...n {
prefixSum[i][j] = prefixSum[i - 1][j] + prefixSum[i][j - 1] - prefixSum[i - 1][j - 1] + matrix[i - 1][j - 1]
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12 changes: 6 additions & 6 deletions lcof2/剑指 Offer II 014. 字符串中的变位词/README.md
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Expand Up @@ -450,29 +450,29 @@ class Solution {
if m > n {
return false
}

var cnt1 = [Int](repeating: 0, count: 26)
var cnt2 = [Int](repeating: 0, count: 26)
let aAscii = Character("a").asciiValue!

for i in 0..<m {
cnt1[Int(s1[s1.index(s1.startIndex, offsetBy: i)].asciiValue! - aAscii)] += 1
cnt2[Int(s2[s2.index(s2.startIndex, offsetBy: i)].asciiValue! - aAscii)] += 1
}

if cnt1 == cnt2 {
return true
}

for i in m..<n {
cnt2[Int(s2[s2.index(s2.startIndex, offsetBy: i)].asciiValue! - aAscii)] += 1
cnt2[Int(s2[s2.index(s2.startIndex, offsetBy: i - m)].asciiValue! - aAscii)] -= 1

if cnt1 == cnt2 {
return true
}
}

return false
}
}
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2 changes: 1 addition & 1 deletion solution/2400-2499/2473.Minimum Cost to Buy Apples/README.md
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Expand Up @@ -27,7 +27,7 @@ tags:

<p>你从某个城市开始，穿越各种道路，最终从&nbsp;<strong>任何一个&nbsp;</strong>城市买&nbsp;<strong>一个&nbsp;</strong>苹果。在你买了那个苹果之后，你必须回到你&nbsp;<strong>开始的&nbsp;</strong>城市，但现在所有道路的成本将&nbsp;<strong>乘以&nbsp;</strong>一个给定的因子 <code>k</code>。</p>

<p>给定整数 <code>k</code>，返回<em>一个大小为 <code>n</code> 的数组 <code>answer</code>，其中 <code>answer[i]</code>&nbsp;是从城市 <code>i</code> 开始购买一个苹果的&nbsp;<strong>最小&nbsp;</strong>总成本。</em></p>
<p>给定整数 <code>k</code>，返回<em>一个大小为 <code>n</code> 的从 1 开始的数组 <code>answer</code>，其中 <code>answer[i]</code>&nbsp;是从城市 <code>i</code> 开始购买一个苹果的&nbsp;<strong>最小&nbsp;</strong>总成本。</em></p>

<p>&nbsp;</p>

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4 changes: 2 additions & 2 deletions solution/3100-3199/3166.Calculate Parking Fees and Duration/README.md
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Expand Up @@ -83,13 +83,13 @@ tags:
<li>从 2023-06-01 10:45:00 到 2023-06-01 12:00:00 在停车场 2：1.25 小时，费用 6.00</li>
<li>从 2023-06-03 07:00:00 到 2023-06-03 09:00:00 在停车场 3：2 小时，费用 4.00</li>
</ul>
总共支付费用：18.00，总小时：7.5，每小时平均费用：2.40，停车场 1 总花费时间最长：2.5 小时。</li>
总共支付费用：18.00，总小时：7.5，每小时平均费用：2.40，停车场 1 总花费时间最长：4.25&nbsp;小时。</li>
<li>对于汽车 ID 1002：
<ul>
<li>从 2023-06-01 09:00:00 到 2023-06-01 11:30:00 在停车场 2：2.5 小时，费用 4.00</li>
<li>从 2023-06-02 12:00:00 到 2023-06-02 14:00:00 在停车场 3：2 小时，费用 2.00</li>
</ul>
总共支付费用：6.00，总小时：4.5，每小时平均费用：1.33，停车场 2 总花费时间最长：4.25 小时。</li>
总共支付费用：6.00，总小时：4.5，每小时平均费用：1.33，停车场 2 总花费时间最长：2.5 小时。</li>
</ul>

<p><b>注意：</b>&nbsp;输出表以 car_id 升序排序。</p>
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2 changes: 1 addition & 1 deletion solution/3100-3199/3166.Calculate Parking Fees and Duration/README_EN.md
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Expand Up @@ -82,7 +82,7 @@ Each row of this table contains the ID of the parking lot, the ID of the car, th
<li>From 2023-06-01 10:45:00 to 2023-06-01 12:00:00 in lot 2: 1.25 hours, fee 6.00</li>
<li>From 2023-06-03 07:00:00 to 2023-06-03 09:00:00 in lot 3: 2 hours, fee 4.00</li>
</ul>
Total fee paid: 18.00, total hours: 7.5, average hourly fee: 2.40, most time spent in lot 1: 2.5 hours.</li>
Total fee paid: 18.00, total hours: 7.5, average hourly fee: 2.40, most time spent in lot 1: 4.25 hours.</li>
<li>For car ID 1002:
<ul>
<li>From 2023-06-01 09:00:00 to 2023-06-01 11:30:00 in lot 2: 2.5 hours, fee 4.00</li>
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4 changes: 2 additions & 2 deletions solution/3100-3199/3167.Better Compression of String/README.md
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Expand Up @@ -38,9 +38,9 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3167.Be

<p><strong>解释：</strong></p>

<p>字符 "a" 和 "b" 在输入中只出现了一次，但 "c" 出现了两次，第一次出现了 9 次，另一次是 1 次。</p>
<p>字符 "a" 和 "b" 在输入中只出现了一次，但 "c" 出现了两次，第一次长度为 9，另一次是长度为 1。</p>

<p>因此，在结果字符串中，它应当出现 10 次。</p>
<p>因此，在结果字符串中，它应当长度为 10。</p>
</div>

<p><strong class="example">示例 2：</strong></p>
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