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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.
求最长回文子串，假设我们已知S[i...j]是回文子串，那么，若S[i-1] = S[j+1]，则S[i-1,...,j+1]是回文子串。
用table[i][j]的true/false表示S[i,...,j]是否为回文子串，
可以得到递推公式：table[i][j] = table[i+1][j-1] && (S[i] == S[j]) (i+1 <= j-1, 即 i+2 <= j)
初始条件：
当j = i 时，table[i][j] = true;
当j = i + 1时，table[i][j] = (S[i] == S[j])
1 string longestPalindrome(string s) 2 { 3 bool table[1000][1000] = {false}; 4 int i, j, slen = s.length(), start = 0, maxlen = 1; 5 6 for (i = 0; i < slen; i++) 7 table[i][i] = true; 8 for (i = 0; i < slen - 1; i++) 9 { 10 j = i + 1; 11 if (s[i] == s[j]) 12 { 13 table[i][j] = true; 14 start = i; 15 maxlen = 2; 16 } 17 } 18 19 for (int len = 3; len <= slen; len++) 20 { 21 for (i = 0; i < slen - len + 1; i++) 22 { 23 j = i + len - 1; 24 if (s[i] == s[j] && table[i + 1][j - 1]) 25 { 26 table[i][j] = true; 27 start = i; 28 maxlen = len; 29 } 30 } 31 } 32 33 return s.substr(start, maxlen); 34 }
leetcode 5.Longest Palindromic Substring
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原文地址：http://www.cnblogs.com/ym65536/p/4207123.html
