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.
最长回文，对于每一个字符从中间向两边找，注意分奇数与偶数两种情况。总的时间复杂度为O(n^2)。另外Manacher算法复杂度为O(n)，Manacher算法好复杂的样子，智商捉急，得好好消化一下。先贴个简单的吧。
1 class Solution { 2 public: 3 int getLPS(const string &s, int idx1, int idx2) { 4 while (idx1 >= 0 && idx2 < s.length() && s[idx1] == s[idx2]) { 5 --idx1; 6 ++idx2; 7 } 8 return idx2 - idx1 - 1; 9 } 10 11 string longestPalindrome(string s) { 12 if (s.length() < 2) return s; 13 int lps = -1, lps1, lps2, pos; 14 for (int i = 0; i < s.length(); ++i) { 15 lps1 = getLPS(s, i, i); 16 lps2 = getLPS(s, i, i + 1); 17 if (lps1 > lps || lps2 > lps) { 18 pos = i; 19 lps = max(lps1, lps2); 20 } 21 } 22 if (lps & 0x1) return s.substr(pos-(lps-1)/2, lps); 23 else return s.substr(pos-(lps-2)/2, lps); 24 } 25 };
[Leetcode] Longest Palindromic Substring
原文：http://www.cnblogs.com/easonliu/p/4379349.html