Longest Palindromic Substring 最长回文子串 学习笔记 1. Brute method 第一种方法:直接循环求解, o(n2) o(n^2) 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution: def longestPalindrome(self, s): """ :type s: str :rtype: str """ l = len(s) max_length = 0 palindromic ...
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_...
2. sol 1 longest common substring问题 3. dp每增加一个字母,最长列长度只可能增加1or2(思路:以当前字母为尾的最长列): class Solution: def longestPalindrome(self, s: str) -> str: if not s: return '' l = 1 start = 0 for i in range(2, len(s)+1): if i>l+1: if s[i-l-2:i...
从中间找到了最后一位或者以最后一个字符为中间key字符的时候结束。这种方法最坏的情况是’ababababababababa……bac’,这种情况的时间复杂度是(0 + 1 + 2 +…+n - 1) = (O(n^2)),由于字符串长度为1000,所以(O(n^2))的时间复杂度是可以接受的。 classSolution(object):deflongestPalindrome(self, s)...
下面是Python的coding代码: classSolution:deflongestPalindrome(self, s):ifnots:return''t ='#'+'#'.join(list(s))+'#'p = [0foriinrange(len(t))] mx =id= resLen = resCenter =0foriinrange(1,len(t)):ifmx > i: p[i] =min(p[2*id- i], mx - i)else: ...
python代码实现class Solution(object): def longestPalindrome(self, s): """ :type s: str :rtype: str """ #动态规划 #ref https://www.cnblogs.com/mini-coconut/p/9074315.html length = len(s) #最长子串开始位置 start = 0 #最长子串长度 --所以最长子串为 s[start,longest] longest = 1 d...
/* * @lc app=leetcode.cn id=5 lang=java * * [5] 最长回文子串 */ // @lc code=start class Solution { public String longestPalindrome(final String s) { if (s.length() == 0) { return s; } String ans; String tmp; ans = s.substring(0, 1); for (int i = 1; i < s.len...
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): ...
class Solution { public: string longestPalindrome(string s) { s.insert(0, "$"); int maxn = 1; int left = 0; int right = 0; for(int i = 0; s[i]; i++){ int left_tmp = i, right_tmp = i; while(s[right_tmp+1] == s[i]) right_tmp++; i = right_tmp; while(s[le...