Leetcode 5. Longest Palindromic Substring

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"

Next challenges: Palindrome Permutation Palindrome Pairs Longest Palindromic Subsequence

方法一：O(n)的时间复杂度。

思路：对原字符串运行"马拉车"算法（Manacher's Algorithm），找到最长回文子串的中心位置所在，然后向两边扩充还原。

关于Manacher's Algorithm的学习资料：

https://segmentfault.com/a/1190000003914228

http://www.cnblogs.com/grandyang/p/4475985.html

代码：

 1 public class Solution {
 2     public String longestPalindrome(String s) {
 3         int maxRight = 0, pos = 0, maxLen = 0, maxPos = 0;
 4         String rs = "#";
 5         for(int i = 0; i < s.length(); i++) rs = rs + s.charAt(i) + "#";
 6         int[] RL = new int[rs.length()];
 7         for(int i = 0; i < rs.length(); i++) {
 8             if(maxRight > i) RL[i] = Math.min(maxRight - i, RL[2 * pos - i]);
 9             while(i - RL[i] - 1>= 0 && i + RL[i] + 1< rs.length() && rs.charAt(i - RL[i] - 1) == rs.charAt(i + RL[i] + 1)) RL[i]++;
10             if(RL[i] + i > maxRight) {
11                 pos = i;
12                 maxRight = RL[i] + i;
13             }
14             if(maxLen < 2 * RL[i] + 1) {
15                 maxLen = 2 * RL[i] + 1;
16                 maxPos = i;
17             }
18         }
19         return rs.substring(maxPos - RL[maxPos], maxPos + RL[maxPos] + 1).replace("#","");
20     }
21 }

方法二：O(n²)的时间复杂度。

思路：dp[i][j]表示s[i,...,j]是否为回文子串。以下2种情况，dp[i][j]为true：

1.j<i+3并且s[i]==s[j]。（这个画图就会明白）

2.j>=i+3并且dp[i+1][j-1]=true。

实现的时候要先保证s[i+1,..,j-1]已经被遍历过。

代码：

 1 public class Solution {
 2     public String longestPalindrome(String s) {
 3         boolean[][] dp = new boolean[s.length()][s.length()];
 4         int maxLen = 0;
 5         String rs = "";
 6         for(int j = 0; j < s.length(); j++) {
 7             int minI = j;
 8             dp[j][j] = true;
 9             for(int i = j - 1; i >= 0; i--) {
10                 if(s.charAt(i) == s.charAt(j) && (j < i + 3 || dp[i + 1][j - 1])) {
11                     dp[i][j] = true;
12                     minI = i;
13                 }else {
14                     dp[i][j] = false;
15                 }
16             }
17             if(j - minI + 1 > maxLen) {
18                 maxLen = j - minI + 1;
19                 rs = s.substring(minI, j + 1);
20             }
21         }
22         return rs;
23     }
24 }