标签：stat pre ++ end asd sdk 记录 开始 panda
题目:最长的回文串
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.
Example 2:
Input: "cbbd"
Output: "bb"
解答：从字符串的第一个元素开始遍历回文串，最短的回文串长度为1，每次只记录最长的回文串。
java实现：
public class LongestPalindromicSubstring { private static int left = 0, length = 0; public static String longestPalindrome(String s) { for(int i = 0; i < s.length(); i++){ expandAroundCenter(s, i); } return s.substring(left, left + length); } private static void expandAroundCenter(String s, int start){ int end = start + 1; start = start - 1; // 巧妙的地方在于如果遇到重复的元素，那么这些重复的元素一定可以作为回文串的中心字符，不需要考虑奇数还是偶数。 while(end < s.length() && (s.charAt(end) == s.charAt(end - 1))) end++; while(start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)){ start--; end++; } if(end - start -1 > length){ left = start + 1; length = end - start -1; } } public static void main(String[] args){ String s = "jfkdslibbasdffdsabbcdfsdkfj"; String result = longestPalindrome(s); System.out.println(result); } }
5. Longest Palindromic Substring
标签：stat pre ++ end asd sdk 记录 开始 panda
原文地址：https://www.cnblogs.com/feng-ying/p/10504597.html
