Longest Palindromic Substring

很经典也很简单的一道题（不考虑非常规解法的话）

下面是中心检索法，时间复杂度O(N^2)

遍历原数组，依次把每个数组元素作为回文的对称中心，向两边搜索

class Solution {public:string longestPalindrome(string s) {int n = s.length();if (n == 0) return "";string longest = s.substr(0, 1);  // a single char itself is a palindrome  for (int i = 0; i < n - 1; i++) {string p1 = expandAroundCenter(s, i, i); //长度为奇数的候选回文字符串  if (p1.length() > longest.length())longest = p1;string p2 = expandAroundCenter(s, i, i + 1);//长度为偶数的候选回文字符串  if (p2.length() > longest.length())longest = p2;}return longest;}string expandAroundCenter(string s, int c1, int c2) {int l = c1, r = c2;int n = s.length();while (l >= 0 && r <= n - 1 && s[l] == s[r]) {l--;r++;}return s.substr(l + 1, r - l-1 );}};