class Solution: def longestPalindrome(self, s: str) -> str: #def longestPalindrome(s: str) -> str: n = len(s) if n < 2: return s # 如果字符串长度小于2,它本身就是最长的回文子串 # dp[i][j]表示s[i:j+1]是否是回文串 dp = [[False] *
* @return */publicstaticStringlongestPalindrome(String s){if(s==null||s.length()<2){returns;}int maxLength=0;String longest=null;int length=s.length();boolean[][]table=newboolean[length][length];// 单个字符都是回文for(int i=0;i<length;i++){table[i][i]=true;longest=s.substring(i...
returns.substring(start, end +1); } privatebooleanisPalindrome(char[] arr,intleft,intright){ while(left < right) { if(arr[left] != arr[right])returnfalse; ++left; --right; } returntrue; } } 『2』中心扩散 解题思路: 首先根据回文串的定义,如果一个字符串正着读和反着读是一样的,那它...
class Solution: def longestPalindrome(self, s: str) -> str: #def longestPalindrome(s: str) -> str: n = len(s) if n < 2: return s # 如果字符串长度小于2,它本身就是最长的回文子串 # dp[i][j]表示s[i:j+1]是否是回文串 dp = [[False] * n for _ in range(n)] start, max_...
class Solution { public: string longestPalindrome(string s) { int len = s.size(); int longest = 0, left = 0, right = 0;//最长长度,左界,右界 vector<vector<bool>> dp(len, vector<bool>(len, false)); for (int i = len - 1; i >= 0; --i) { for (int j = i; j < le...
参考:LeetCode:Longest Palindromic Substring 最长回文子串 - tenos中的方法4 动态规划 AC代码: 代码语言:javascript 代码运行次数:0 classSolution{public:stringlongestPalindrome(string s){constint len=s.size();if(len<=1)returns;bool dp[len][len];//dp[i][j]表示s[i..j]是否是回文memset(dp,0,...
def longestPalindrome(self, s: str) -> str: l, ms = len(s), s[0] if s else '' # 内循环少一次 if l < 2: return s for i in range(l): for j in range(i + len(ms), l):# 关键点一: + len(ms) x = s[i:j + 1] ...
Can you solve this real interview question? Longest Palindromic Substring - Given a string s, return the longest palindromic substring in s. Example 1: Input: s = "babad" Output: "bab" Explanation: "aba" is also a valid answer. Example 2: Input
public String longestPalindrome(String s) { // 改造字符串,每个字符间添加#。添加头^尾$两个不同的字符用于消除边界判断 StringBuilder sb = new StringBuilder("^"); for (int i = 0, len = s.length(); i < len; i++) sb.append("#").append(s.charAt(i)); sb.append("#$"); int c ...
(2)中心扩展法。复杂度O(N²) 枚举每个字符作为中心点向左右扩展。 可是这里要注意,对于每一次扩展要分奇偶两种情况。 否则可能会漏掉情况。 一发现不满足的情况立即break进行下一个中心字符的推断,代码例如以下: class Solution { public: string longestPalindrome(string s) { string ans; int len=s.length...