[Swift]LeetCode647. 回文子串 | Palindromic Substrings
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:https://www.cnblogs.com/strengthen/p/10482491.html
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
Given a string, your task is to count how many palindromic substrings in this string.
The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.
Example 1:
Input: "abc" Output: 3 Explanation: Three palindromic strings: "a", "b", "c".
Example 2:
Input: "aaa" Output: 6 Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
Note:
- The input string length won't exceed 1000.
给定一个字符串,你的任务是计算这个字符串中有多少个回文子串。
具有不同开始位置或结束位置的子串,即使是由相同的字符组成,也会被计为是不同的子串。
示例 1:
输入: "abc" 输出: 3 解释: 三个回文子串: "a", "b", "c".
示例 2:
输入: "aaa" 输出: 6 说明: 6个回文子串: "a", "a", "a", "aa", "aa", "aaa".
注意:
- 输入的字符串长度不会超过1000。
16ms
1 class Solution { 2 func countSubstrings(_ s: String) -> Int { 3 var count = 0 4 var s = Array(s) 5 for i in 0..<s.count { 6 if i + 1 < s.count, s[i] == s[i + 1] { 7 count += numPalindromes(s, i, i + 1) 8 } 9 count += numPalindromes(s, i, i) 10 } 11 return count 12 } 13 14 func numPalindromes(_ s: [Character], _ i: Int, _ j: Int) -> Int { 15 var count = 0, i = i, j = j 16 while i >= 0, j < s.count, s[i] == s[j] { 17 i -= 1 18 j += 1 19 count += 1 20 } 21 return count 22 } 23 }
20ms
1 class Solution { 2 func countSubstrings(_ s: String) -> Int { 3 var result = 0 4 let input = Array(s) 5 for i in 0..<input.count { 6 checkPalindrom(i,i,input,&result) 7 checkPalindrom(i,i + 1,input,&result) 8 } 9 return result 10 } 11 12 func checkPalindrom(_ left: Int, _ right: Int, _ input: [Character], _ result: inout Int) { 13 let count = input.count 14 var i = left, j = right 15 while i >= 0, j < count { 16 if input[i] == input[j] { 17 result += 1 18 i -= 1 19 j += 1 20 }else{ 21 break 22 } 23 } 24 } 25 }
28ms
1 class Solution { 2 func countSubstrings(_ s: String) -> Int { 3 let arr = Array(s) 4 let cnt = s.count 5 if cnt == 0 { 6 return 0 7 } 8 var result = 0 9 10 for center in 0..<cnt { 11 var left = center 12 var right = center 13 14 while left >= 0 && right < cnt && arr[left] == arr[right] { 15 result += 1 16 left -= 1 17 right += 1 18 } 19 } 20 21 for center in 1..<cnt { 22 var left = center-1 23 var right = center 24 25 while left >= 0 && right < cnt && arr[left] == arr[right] { 26 result += 1 27 left -= 1 28 right += 1 29 } 30 } 31 return result 32 } 33 }
32ms
1 class Solution { 2 func countSubstrings(_ s: String) -> Int { 3 let s = Array(s) 4 var n = s.count, ans = 0 5 for center in 0..<2*n { 6 var left = center / 2 7 var right = left + center % 2 8 while (left >= 0 && right < n && 9 s[s.index(s.startIndex,offsetBy:left)] == 10 s[s.index(s.startIndex,offsetBy:right)]) { 11 ans+=1 12 left-=1 13 right+=1 14 } 15 } 16 return ans 17 } 18 }
40ms
1 class Solution { 2 func countSubstrings(_ s: String) -> Int { 3 let chars = Array(s) 4 func extend(_ left: Int, _ right: Int) -> Int { 5 var (left, right) = (left, right) 6 var count = 0 7 while left >= 0 8 && right < chars.count 9 && chars[left] == chars[right] { 10 left -= 1 11 right += 1 12 count += 1 13 } 14 15 return count 16 } 17 18 return chars.indices.map { 19 extend($0, $0) + extend($0, $0 + 1) 20 } 21 .reduce(0, +) 22 } 23 }
56ms
1 class Solution { 2 func countSubstrings(_ s: String) -> Int { 3 guard s.count > 1 else{ 4 return 1 5 } 6 7 var manipulationStr : [Character] = [] 8 for char in Array(s){ 9 manipulationStr.append("#") 10 manipulationStr.append(char) 11 } 12 manipulationStr.append("#") 13 14 var count : Int = 0 15 for movingIndex in 1..<manipulationStr.count - 1{ 16 var leftIndex = movingIndex 17 var rightIndex = movingIndex 18 while leftIndex >= 0 && rightIndex < manipulationStr.count{ 19 if manipulationStr[leftIndex] != manipulationStr[rightIndex]{ 20 break 21 }else{ 22 if manipulationStr[leftIndex] != "#"{ 23 count += 1 24 } 25 leftIndex -= 1 26 rightIndex += 1 27 } 28 } 29 } 30 return count 31 } 32 }
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
· 没有源码,如何修改代码逻辑?
· 全程不用写代码,我用AI程序员写了一个飞机大战
· MongoDB 8.0这个新功能碉堡了,比商业数据库还牛
· DeepSeek 开源周回顾「GitHub 热点速览」
· 记一次.NET内存居高不下排查解决与启示
· 白话解读 Dapr 1.15:你的「微服务管家」又秀新绝活了