标签：
https://oj.leetcode.com/problems/longest-palindromic-substring/
Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.
一种比较朴素的解法，从左至右遍历数组，以当前数字为轴，或者以该数字组成的两个相同数字为轴，（两种情况），向两边扩散。终止条件为，扩散到数组开始或结尾，或者两段元素不一致。记录每种情况的长度，最后返回最长的开始和结束坐标。该解法主要就是注意两种情况。时间复杂度为O(n^2)。
public class Solution { public String longestPalindrome(String s) { int start = 0; int end = 0; int maxStart = 0; int maxEnd = 0; int maxLength = 0; //以某数为对称轴 for(int i = 0; i < s.length(); i++){ start = i; end = i; while(start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)){ start--; end++; } start++; end--; int length = end - start + 1; if(length > maxLength){ maxLength = length; maxStart = start; maxEnd = end; } } //以某两个对称 if(s.length() > 1){ for(int i = 1; i < s.length(); i++){ if(s.charAt(i) == s.charAt(i - 1)){ start = i -1; end = i; while(start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)){ start--; end++; } } start++; end--; int length = end - start + 1; if(length > maxLength){ maxLength = length; maxStart = start; maxEnd = end; } } } return s.substring(maxStart, maxEnd + 1); } }
标签：
原文地址：http://www.cnblogs.com/NickyYe/p/4234832.html
