Returns the maximum number of palindromes in a string

/*** @author pshdhx* @date 2022-08-03 15:20* @Des gives you a string s, find the longest palindrome substring in s.* input: s = "babad"* output: "bab"* Explanation: "aba" is also an answer that matches the meaning of the question.* 
* input: s = "cbbd"* output: "bb"* @Method* If a string is a palindrome, then removing its first and last letters is still a palindrome;* @Summary*/

package com.pshdhx.Algorithm.middle;/*** @author pshdhx* @date 2022-08-03 15:20* @Des gives you a string s, find the longest palindrome substring in s.* input: s = "babad"* output: "bab"* Explanation: "aba" is also an answer that matches the meaning of the question.* 
* input: s = "cbbd"* output: "bb"* @Method* If a string is a palindrome, then removing its first and last letters is still a palindrome;* @Summary*/public class longest palindrome substring {public String longestPalindrome(String s) {int len ​​= s.length();if (len < 2) {return s;}int maxLen = 1;int begin = 0;// dp[i][j] indicates whether s[i..j] is a palindromeboolean[][] dp = new boolean[len][len];// initialization: all substrings of length 1 are palindromesfor (int i = 0; i < len; i++) {dp[i][i] = true;}char[] charArray = s.toCharArray();// start recursionfor (int L = 2; L <= len; L++) {// length of palindromefor (int i = 0; i < len; i++) {//left boundary of palindrome// The right boundary can be determined by L and i, i.e. j - i + 1 = L we getint j = L + i - 1;//The right boundary of the palindrome// If the right boundary is out of bounds, you can exit the current loop [Note: the right boundary starts from 0, it cannot be greater than or equal to the length of the array]if (j >= len) {break;}if (charArray[i] != charArray[j]) {//If left boundary != right boundarydp[i][j] = false;} else {// left border == right borderif (j - i < 3) {//The length of the palindrome string is less than 4====" [1,2,3] If it is 2: left border = right border, true; if it is 3, then left border = right border, true, regardless of what is in the middle!dp[i][j] = true;} else {dp[i][j] = dp[i + 1][j - 1];}}// As long as dp[i][L] == true, it means that the substring s[i..L] is a palindrome, and record the length and starting position of the palindromeif (dp[i][j] && j - i + 1 > maxLen) {maxLen = j - i + 1;begin = i;}}}return s.substring(begin, begin + maxLen);}public static void main(String[] args) {String babad = new longest palindrome().longestPalindrome("babad");System.out.println(babad);}}