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题目链接:https://oj.leetcode.com/problems/longest-palindromic-substring/
这道题通常可以写出来的是2种做法。
1. 保持一个index从前往后扫,每次index的循环中,保持left和right的index向两边扫,但是有2种情况,aba和abba。left和right的扫描要扫2次
2. DP做法,2维矩阵
* 可以适当减枝,比如当前index完全不可能提供比已经得到的最长string再长的可能,则移动到下一个index
class Solution { public: string longestPalindrome(string s) { if (s.size() == 0) return ""; int idx = 0; int maxLen = 0; string res; while (idx < s.size()) { // break if even the longest palindrome string we could get // is still shorter than the maximun one we have if (min(s.size() - idx, idx + 1) * 2 <= maxLen) break; int left; int right; // 1st case, ex: aba left = idx; right = idx; while (left >= 0 && right < s.size()) { if (s[left] == s[right]) { if (maxLen < right - left + 1) { maxLen = right - left + 1; res = s.substr(left, right - left + 1); } --left; ++right; } else { break; } } // 2nd case, ex: abba left = idx; right = idx + 1; while (left >= 0 && right < s.size()) { if (s[left] == s[right]) { if (maxLen < right - left + 1) { maxLen = right - left + 1; res = s.substr(left, right - left + 1); } --left; ++right; } else { break; } } ++idx; } return res; } };
[Leetcode] 5 - Longest Palindromic Substring
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原文地址:http://blog.csdn.net/algorithmengine/article/details/42492003