5. Longest Palindromic Substring（求最长回文子字符串）

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"Output: "bab"Note: "aba" is also a valid answer.

Example:

Input: "cbbd"Output: "bb"

题目大意：给定一个字符串，找到最长的回文子串。

解法一：

解题思路：回文字符串有两种情况：长度为奇数，或者长度为偶数。现在分别以这两种情况进行解题，两个int类型的指针left和right，分别向两侧延伸，以找到长度最大的回文子字符串。

解题代码：（44ms，beats 44.34%）

class Solution {    public String longestPalindrome(String s) {        if (s.length() <= 1) {return s;}int left, right;String res = s.substring(0, 1);for (int i = 0; i < s.length(); i++) {left = right = i;while (left > -1 && right < s.length() && s.charAt(left) == s.charAt(right)) {left--;right++;}if (right - left - 1 > res.length()) {res = s.substring(left + 1, right);}}for (int i = 0; i < s.length() - 1; i++) {left = i;right = i + 1;while (left > -1 && right < s.length() && s.charAt(left) == s.charAt(right)) {left--;right++;}if (right - left - 1 > res.length()) {res = s.substring(left + 1, right);}}return res;    }}

解法二：

解题思路：动态规划

解题代码：（180ms，beats 1.67%）

class Solution {    public String longestPalindrome(String s) {        int sLen = s.length();if (sLen <= 1) {return s;}boolean[][] isPalindrome = new boolean[sLen][sLen];// 初始化，i=j时只有一个字符，是回文字符串。i>j时是空串，也视为回文字符串。i<j时先默认初始化为false。for (int i = 0; i < sLen; i++) {for (int j = 0; j < sLen; j++) {if (i < j) {isPalindrome[i][j] = false;} else {isPalindrome[i][j] = true;}}}int left = 0, right = 0;for (int step = 1; step < sLen; step++) {for (int i = 0; i + step < sLen; i++) {int j = i + step;if (s.charAt(i) != s.charAt(j)) {isPalindrome[i][j] = false;} else {isPalindrome[i][j] = isPalindrome[i + 1][j - 1];if (isPalindrome[i][j] == true) {if (j - i > right - left) {left = i;right = j;}}}}}return s.substring(left, right + 1);    }}