Longest Palindromic Substring
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简单的动态规划,一个子串两头相等的话,它是不是回文的就取决与掐掉两头后剩下的部分是不是回文的。
网上有O(n)的解法,不过等我有时间了再看吧。
class Solution {public: bool DP[1001][1001]; string longestPalindrome(string s) { memset(DP, false, sizeof(DP)); for (int i = 0; i < s.length(); ++i) DP[i][i] = true; int start, end; for (int len = 2; len <= s.length(); ++len) { for (int i = 0; len + i - 1 < s.length(); ++i) { int j = len + i - 1; if (s[i] != s[j]) DP[i][j] = false; else { if (j == i + 1) DP[i][j] = true; else DP[i][j] = DP[i + 1][j - 1]; if (DP[i][j]) { start = i; end = j; } } } } return s.substr(start, end - start + 1); }};
http://oj.leetcode.com/problems/longest-palindromic-substring/
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