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.
本文利用动态规划的思想，如果s[i]==s[j]，那么s[i+1]==s[j-1]时，才是子串。时间复杂度O(n2)。
代码如下：
string longestPalindrome(string s) { int n = s.size(); int substrBegin = 0; int maxlen = 1; bool state[1000][1000] = { false }; for (int i = 0; i < n; i++){ state[i][i] = true; if (i < n - 1 && s[i] == s[i + 1]){ state[i][i + 1] = true; substrBegin = i; maxlen = 2; } } for (int i = 3; i <= n ; i++){ for (int j = 0; j < n - i + 1; j++){ if (s[j] == s[j + i - 1] && state[j + 1] [j+i-2]== true){ state[j][j+i - 1] = true; substrBegin = j; maxlen = i; } } } return s.substr(substrBegin, maxlen); }
#5 Longest Palindromic Substring
原文：http://www.cnblogs.com/Scorpio989/p/4399328.html