LeetCode之Longest Palindromic Substring

/*用dp[i][j]表示子串s[i...j]是否为回文，则有
dp[i][j] = d[i+1][j-1] && s[i] == s[j](i+1 < j);
dp[i][j] = (s[i] == s[j])(i+1 == j)并且，dp[i][i] = true;。*/
class Solution {
public:
	string longestPalindrome(string s) {
		if(s.size() <= 1) return s;
		int len = 1, start = 0;
		const int size = s.size();
		bool dp[size][size];
		memset(dp, 0, sizeof(dp));
		for(int i = 0; i < s.size(); ++i) dp[i][i] = true;
		for(int j = 1; j < s.size(); ++j){
			for(int i = 0; i + j < s.size(); ++i){
				if(j == 1) dp[i][i+j] = (s[i] == s[i+j]);
				else dp[i][i+j] = (dp[i+1][i+j-1] && (s[i] == s[i+j]));
				if(dp[i][i+j] && len < j+1){
					len = j+1;
					start = i;
				}
			}
		}
		return s.substr(start, len);
	}
};