mirror of https://github.com/doocs/leetcode.git
feat: update lc problems (#2988)
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Libin YANG
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@ -188,13 +188,13 @@ class Solution {
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func maxProduct(_ words: [String]) -> Int {
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let n = words.count
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var masks = [Int](repeating: 0, count: n)
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for i in 0..<n {
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for c in words[i] {
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masks[i] |= 1 << (c.asciiValue! - Character("a").asciiValue!)
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}
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}
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var maxProduct = 0
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for i in 0..<n {
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for j in i+1..<n {
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@ -203,7 +203,7 @@ class Solution {
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}
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}
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}
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return maxProduct
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}
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}
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@ -185,7 +185,7 @@ class Solution {
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var ans = inf
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var sum = 0
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var i = 0
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for j in 0..<nums.count {
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sum += nums[j]
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while sum >= target {
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@ -194,7 +194,7 @@ class Solution {
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i += 1
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}
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}
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return ans == inf ? 0 : ans
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}
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}
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@ -155,11 +155,11 @@ function numSubarrayProductLessThanK(nums: number[], k: number): number {
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class Solution {
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func numSubarrayProductLessThanK(_ nums: [Int], _ k: Int) -> Int {
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if k <= 1 { return 0 }
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var product: Int = 1
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var ans: Int = 0
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var left: Int = 0
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for right in 0..<nums.count {
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product *= nums[right]
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while product >= k {
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@ -168,7 +168,7 @@ class Solution {
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}
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ans += right - left + 1
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}
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return ans
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}
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}
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@ -151,13 +151,13 @@ class Solution {
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var cnt: [Int: Int] = [0: 1]
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var ans = 0
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var s = 0
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for x in nums {
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s += x
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ans += cnt[s - k, default: 0]
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cnt[s, default: 0] += 1
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}
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return ans
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}
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}
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@ -169,7 +169,7 @@ class Solution {
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var d: [Int: Int] = [0: -1]
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var ans = 0
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var s = 0
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for i in 0..<nums.count {
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s += nums[i] == 0 ? -1 : 1
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if let prevIndex = d[s] {
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@ -178,7 +178,7 @@ class Solution {
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d[s] = i
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}
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}
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return ans
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}
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}
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@ -233,7 +233,7 @@ class Solution {
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func pivotIndex(_ nums: [Int]) -> Int {
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var leftSum = 0
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var rightSum = nums.reduce(0, +)
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for i in 0..<nums.count {
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rightSum -= nums[i]
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if leftSum == rightSum {
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@ -264,7 +264,7 @@ class NumMatrix {
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let m = matrix.count
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let n = matrix[0].count
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prefixSum = Array(repeating: Array(repeating: 0, count: n + 1), count: m + 1)
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for i in 1...m {
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for j in 1...n {
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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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@ -450,29 +450,29 @@ class Solution {
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if m > n {
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return false
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}
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var cnt1 = [Int](repeating: 0, count: 26)
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var cnt2 = [Int](repeating: 0, count: 26)
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let aAscii = Character("a").asciiValue!
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for i in 0..<m {
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cnt1[Int(s1[s1.index(s1.startIndex, offsetBy: i)].asciiValue! - aAscii)] += 1
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cnt2[Int(s2[s2.index(s2.startIndex, offsetBy: i)].asciiValue! - aAscii)] += 1
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}
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if cnt1 == cnt2 {
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return true
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}
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for i in m..<n {
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cnt2[Int(s2[s2.index(s2.startIndex, offsetBy: i)].asciiValue! - aAscii)] += 1
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cnt2[Int(s2[s2.index(s2.startIndex, offsetBy: i - m)].asciiValue! - aAscii)] -= 1
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if cnt1 == cnt2 {
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return true
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}
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}
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return false
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}
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}
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@ -27,7 +27,7 @@ tags:
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<p>你从某个城市开始，穿越各种道路，最终从 <strong>任何一个 </strong>城市买 <strong>一个 </strong>苹果。在你买了那个苹果之后，你必须回到你 <strong>开始的 </strong>城市，但现在所有道路的成本将 <strong>乘以 </strong>一个给定的因子 <code>k</code>。</p>
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<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> 是从城市 <code>i</code> 开始购买一个苹果的 <strong>最小 </strong>总成本。</em></p>
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<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> 是从城市 <code>i</code> 开始购买一个苹果的 <strong>最小 </strong>总成本。</em></p>
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<p> </p>
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@ -83,13 +83,13 @@ tags:
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<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>
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<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>
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</ul>
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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>
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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 小时。</li>
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<li>对于汽车 ID 1002：
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<ul>
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<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>
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<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>
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</ul>
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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>
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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>
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</ul>
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<p><b>注意：</b> 输出表以 car_id 升序排序。</p>
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@ -82,7 +82,7 @@ Each row of this table contains the ID of the parking lot, the ID of the car, th
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<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>
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<li>From 2023-06-03 07:00:00 to 2023-06-03 09:00:00 in lot 3: 2 hours, fee 4.00</li>
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</ul>
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Total fee paid: 18.00, total hours: 7.5, average hourly fee: 2.40, most time spent in lot 1: 2.5 hours.</li>
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Total fee paid: 18.00, total hours: 7.5, average hourly fee: 2.40, most time spent in lot 1: 4.25 hours.</li>
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<li>For car ID 1002:
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<ul>
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<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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@ -38,9 +38,9 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3167.Be
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<p><strong>解释：</strong></p>
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<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>
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<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>
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<p>因此，在结果字符串中，它应当出现 10 次。</p>
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<p>因此，在结果字符串中，它应当长度为 10。</p>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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