LeetCode-Longest Palindromic Substring（JAVA）

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

Example 1:

Input: “babad” Output: “bab” Note: “aba” is also a valid answer.
Example 2:

Input: “cbbd” Output: “bb”

求給定字符串s的最長迴文子串

1. 暴力破解

在第i個字符串時，比較i-k與i+k是否相等

2. dp

dp[i][j] 爲字符串的第i到第j個字符串是否爲迴文字符串

• 遞推公式：dp[i][j] = (s.charAt(i) == s.charAt(j) && (dp[i+1][j - 1] || j-i<=2))

dp數組的初始化爲dp[i][i] =1

class Solution {
    public String longestPalindrome(String s) {
      int n = s.length();
        if(n < 2) {
            return s;
        }
        
        // 定義一個數組dp[n][n],dp[i][j]標識索引i和j之間是否是迴文子序列
        //遞推公式：dp[i][j] = if((s.charAt(i) == s.charAt(j)) && (dp[i+1][j-1] || i-j <=2 )) 
        
        
        boolean[][] dp = new boolean[n][n];
        for(int i = 0; i< n ; i++) {
            dp[i][i] = true;
        }
        
        int l = 0;
        int r = 0;
        int len = 1;
        for(int j = 1; j < n; j++) {
            for(int i = 0; i< j ; i++) {
                if(s.charAt(i) == s.charAt(j) &&(dp[i + 1][j - 1] || j - i <= 2)) {
                    dp[i][j] = true;
                    int len1 = j - i + 1;
                    if(len1 > len) {
                        l = i;
                        r = j;
                        len = len1;
                    }
                } else {
                    dp[i][j] = false;
                }
            }
        }
        return s.substring(l, r + 1);
    }
}