leetcode5最长回文串

Posted 2021-09-10 lisin-lee-cooper

一.问题描述

给你一个字符串 s，找到 s 中最长的回文子串。

示例 1：

输入：s = “babad”
输出：“bab”
解释：“aba” 同样是符合题意的答案。

二.示例代码

public class LongestPalindromeStr5 {

    public static void main(String[] args) {

        String str = "babad";
        String result = longestPalindromeStr2(str);
        System.out.println(result);
    }

    private static String longestPalindromeStr(String str) {

        String result = "";
        int maxLen = 0;
        for (int i = 1; i < str.length(); i++) {
            for (int j = 0; j < i; j++) {
                if (isPalindromeStr(str.substring(j, i)) && i - j > maxLen) {
                    maxLen = i - j;
                    result = str.substring(j, i);
                }
            }
        }
        return result;
    }

    private static boolean isPalindromeStr(String palindromeStr) {
        for (int i = 0; i < palindromeStr.length() / 2; i++) {
            if (palindromeStr.charAt(i) != palindromeStr.charAt(palindromeStr.length() - i - 1)) {
                return false;
            }
        }
        return true;
    }

    public static String longestPalindromeStr2(String s) {
        int strLen = s.length();
        if (strLen < 2) {
            return s;
        }
        boolean[][] dp = new boolean[strLen][strLen];
        for (int i = 0; i < strLen; i++) {
            dp[i][i] = true;
        }

        int begin = 0;
        int maxLen = 0;
        for (int j = 1; j < strLen; j++) {
            for (int i = 0; i < j; i++) {
                if (s.charAt(i) != s.charAt(j)) {
                    dp[i][j] = false;
                } else {
                    if (j - i < 3) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                }

                if (dp[i][j] && j - i + 1 > maxLen) {
                    maxLen = j - i + 1;
                    begin = i;
                }
            }
        }
        return s.substring(begin, begin + maxLen);
    }


}