思路一实现: class Solution { public: string longestPalindrome(string s) { string res = s.substr(0,1); int length = 1; bool ** P = new bool*[s.length()]; for(int i = 0; i < s.length(); i++) { P[i] = new bool[s.length()]; P[i][i] = true; if(i != s.length...
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...
public class Solution { /** * @param s: input string * @return: the longest palindromic substring */ public String longestPalindrome(String s) { // write your code here char[] S = s.toCharArray(); int sLength = S.length; String[][] f = new String[sLength ][sLength ]; return ...
Given a strings, returnthe longestpalindromicsubstringins. Example 1: Input:s = "babad"Output:"bab"Explanation:"aba" is also a valid answer. Example 2: Input:s = "cbbd"Output:"bb" Constraints: 1 <= s.length <= 1000 sconsist of only digits and English letters. Accepted 3.8M Submission...
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 ...
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] ...
def longestPalindrome(self, s): ans = '' max_len = 0 n = len(s) DP = [[False] * n for _ in range(n)] # 字符串长度为1 for i in range(n): DP[i][i] = True max_len = 1 ans = s[i] # 字符串长度为2 for i in range(n-1): ...
在LeetCode上本题属于Medium难度。是典型的动态规划类型题目:直接上代码,代码中有详细的注释;代码中提供了2种解法,暴力解法和动态规划 class Solution { public String longestPalindrome(String s){ int length = s.length(); boolean[][] dp = new boolean[length][length]; int startIndex = 0; int maxLen...
LeetCode刷题，是为了获得面经，这是题目5的java实现解法 工具/原料 笔记本 eclipse 方法/步骤 1 题目叙述Given a stringS, find the longest palindromic substring inS. You may assume that the maximum length ofSis 1000, and there exists one unique longest palindromic substring.2 本体可以使用动态规划去...
publicStringlongestPalindrome(Strings){Stringresult="";char[]characters=s.toCharArray();for(inti=0;i<s.length();i++){StringtmpStr=s.substring(i);if(result.length()>=tmpStr.length())returnresult;while(tmpStr.length()>0){booleanisReverse=isReverseMethod(tmpStr);if(isReverse){//是回文if(...