Leetcode 5 longest palindrome substring

Give you a string s and find the longest palindrome substring in s

Example 1:

Input: s = "babad"
Output: "bab"
Explanation: "aba" is also an answer that meets the meaning of the question.

Example 2:

Input: s = "cbbd"
Output: "bb"

Example 3:

Input: s = "a"
Output: "a"

Example 4:

Input: s = "ac"
Output: "a"

Problem-solving ideas

Using dynamic programming, the state transition equation is as follows:

dp[i][j] = (s[i] == s[j]) && (i - j < 3 || dp[j + 1][i - 1])

If s[i] == s[j] appears, then if i - j < 3 or dp[j + 1][i - 1] is also a palindrome string, then the string from j to i is also a palindrome string.

i - j < 3 corresponds to three situations, the first is three characters, the beginning and the end are the same, no matter what character is in the middle, it is a palindrome, the second is two characters, both are the same, obviously also a palindrome, and the third is There is only one character, and a character itself is also a palindrome

Java implementation:

class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0) return s;
        int len = s.length();
        int start = 0;
        int maxLen = 1;
        boolean[][] dp = new boolean[len][len];
        for (int i=0; i<len; i++) {
            for (int j=0; j<=i; j++) {
                if (i == j) {
                    // 初始条件
                    // 一个字符本身就是回文串
                    dp[i][i] = true;
                    continue;
                }
                if (s.charAt(i) == s.charAt(j)) {
                    if (i - j <= 2 || dp[j + 1][i - 1]) {
                        // 状态转移方程
                        dp[j][i] = true;
                        if (i - j + 1 > maxLen) {
                            // 记录最大长度和起始下标
                            maxLen = i - j + 1;
                            start = j;
                        }
                    }
                }
            }
        }
        // 返回字符串
        return s.substring(start, start + maxLen);
    }
}

JS implementation:

function longestPalindrome(s) {
    if (s == null || s.length == 0) return s;
    let len = s.length;
    let start = 0;
    let maxLen = 1;
    let dp = gen2DArray(len, len);
    for (let i=0; i<s.length; i++) {
        for (let j=0; j<=i; j++) {
            if (i == j) {
                dp[i][i] = true;
                continue;
            }
            if (s[i] == s[j]) {
                if (i - j <= 2 || dp[j + 1][i - 1]) {
                    dp[j][i] = true;
                    if (i - j + 1 > maxLen) {
                        maxLen = i - j + 1;
                        start = j;
                    }
                }
            }
        }
    }
    return s.substring(start, start + maxLen)
}

function gen2DArray(a, b) {
    let res = [];
    for (let i=0; i<a; i++) {
        res.push([]);
        for (let j=0; j<b; j++) {
            res[i][j] = false;
        }
    }
    return res;
}