Force deduction 5: Longest palindrome substring

subject ： Palindrome string （ Left to right and right to left strings are the same ）

Ideas ：

1. Suppose the length of the string is 1;

2. Set a two-dimensional array of boolean type , Using two Pointers i,j Traverse them separately i=0,j<=i;

3. Judge , Use charAt() Method to judge , Whether two strings are equal , And whether their length is equal to 2

Code ：

class Solution {
    public String longestPalindrome(String s) {
        //  Special case judgment 
        if(s.length() < 2 || s == null) return s;

        int len = s.length(); //  Length of string 
        int start = 0; //  The starting point of the longest palindrome substring 
        int end = 0; //  The end of the longest palindrome substring 
        int max = 1; //  The length of the longest palindrome substring 

        boolean[][] dp = new boolean[len][len];//  Create a 2D array , For storage status 

        for(int i = 0; i < len; i++){
            for(int j = 0; j <= i; j++){
                if(s.charAt(j) == s.charAt(i) && (i-j < 2 || dp[j+1][i-1])){
                    dp[j][i] = true;
                    if(i-j+1 > max){
                        start = j;
                        end = i;
                        max = Math.max(max, i-j+1);
                    }
                }
            }
        }

        return s.substring(start, end+1);
    }
}

details ：

charAt() Method to return the characters at the specified index . Index range is from 0 To length() - 1.

Reference resources ：

(97 Bar message ) 【Java Language 】 Force buckle series ----5. Longest text substring _ A blog about Xiaobai's struggle -CSDN Blog

(97 Bar message ) Force button brush questions ： Longest text substring （java Realization ）_ Qianqianjun's blog -CSDN Blog _ Longest text substring java