不管怎么说,每天还是要水一篇博文,写一个LeetCode的题才圆满。 题目:Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000. Example:Input: "babad" ; Output: "bab"; Note: "aba
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
【LeetCode】5. Longest Palindromic Substring 陌儿的百科全书 来自专栏 · LeetCode 题目: Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000. Example 1: Input: "babad" Output: "bab" Note: "aba" is also a valid answer. ...
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. https://oj.leetcode.com/problems/longest-palindromic-substring/ 思路1(naive approach):依次检查所有的子串(n^2),判断是否是palin...
情况1:以i'为中心的最长回文串完全包含在L-R之内,那么T和T'也是关于C点镜面对称的,无需再对i进行扩展。 情况2:以i'为中心的最长回文串超过了左端点L,那么该串的中间部分T'和以i为中心的回文串的中间部分T是关于C点镜面对称的,之后在T的基础上继续向两边扩展即可。
LeetCode算法题有一个技巧:先看每道题的example,大部分情况看完example就能理解题目意思;如果看完example还有疑惑,可以再回过头来看英文题目。这样也可以节省一点时间~ 题目描述 Given a string s, return the longest palindromic substring in s. 经典的题目,最长回文子串,所谓回文字符串:正反字符串相等 Examples1th...
LeetCode第5道题目:5. Longest Palindromic Substring Given a strings, find the longest palindromic substring ins. You may assume that the maximum length ofsis 1000. Example 1: Input: "babad" Output: "bab" Note: "aba" is also a valid answer. ...
leetcode 5. Longest Palindromic Substring,给一个字符串求最长公共字串。直olution(object):defpre(self,s):lenS=len(s)i=0res=[]whilei<lenS:res.append('
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...
{ C = i; R = i + P[i]; } } // Find the maximum element in P. int maxLen = 0; int centerIndex = 0; for (int i = 1; i < n-1; i++) { if (P[i] > maxLen) { maxLen = P[i]; centerIndex = i; } } delete[] P; return s.substr((centerIndex - 1 - maxLen...