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"

 

Similar Questions: Shortest Palindrome Palindrome Permutation Palindrome Pairs Longest Palindromic Subsequence Palindromic Substrings 

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：

public class Solution {
    public String longestPalindrome(String s) {
        int maxRight = 0, pos = 0, maxLen = 0, maxPos = 0;
        String rs = "#";
        for(int i = 0; i < s.length(); i++) rs = rs + s.charAt(i) + "#";
        int[] RL = new int[rs.length()];
        for(int i = 0; i < rs.length(); i++) {
            if(maxRight > i) RL[i] = Math.min(maxRight - i, RL[2 * pos - i]);
            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]++;
            if(RL[i] + i > maxRight) {
                pos = i;
                maxRight = RL[i] + i;
            }
            if(maxLen < 2 * RL[i] + 1) {
                maxLen = 2 * RL[i] + 1;
                maxPos = i;
            }
        }
        return rs.substring(maxPos - RL[maxPos], maxPos + RL[maxPos] + 1).replace("#","");
    }
}

 

 

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：

public class Solution {
    public String longestPalindrome(String s) {
        boolean[][] dp = new boolean[s.length()][s.length()];
        int maxLen = 0;
        String rs = "";
        for(int j = 0; j < s.length(); j++) {
            int minI = j;
            dp[j][j] = true;
            for(int i = j - 1; i >= 0; i--) {
                if(s.charAt(i) == s.charAt(j) && (j < i + 3 || dp[i + 1][j - 1])) {
                    dp[i][j] = true;
                    minI = i;
                }else {
                    dp[i][j] = false;
                }
            }
            if(j - minI + 1 > maxLen) {
                maxLen = j - minI + 1;
                rs = s.substring(minI, j + 1);
            }
        }
        return rs;
    }
}