原题传送门

1. 问题描述

2. Solution 1: Brute Force

1、思路
两层循环遍历所有子串，判断是否为回文，保留最长的子串。
2、代码实现

import sys

if sys.platform != "linux":
    file_in = open("input/HJ85.txt")
    sys.stdin = file_in


def check(s, i, j):
    start = i
    end = j
    while start <= end:
        if s[start] != s[end]:
            return False
        start += 1
        end -= 1
    return True


def solve(s):
    max_len = 1
    for i in range(len(s)):
        for j in range(i + 1, len(s)):
            if check(s, i, j) and (j - i + 1) > max_len:
                max_len = j - i + 1
    print(max_len)


for line in sys.stdin:
    s = line.strip()
    solve(s)

3. Solution 2

1、思路： 用DP优化暴力法
Dynamic Programming

1. 状态定义
   boolean dp[i][j] = s[i, j] is Palindromic
2. 状态初始条件
   dp[i][j] = true if len(s[i, j]) == 1 or j == i or j - i == 0;
   dp[i][j] = s[i] == s[j] if len(s[i, j]) == 2 or j == (i + 1) or j - i == 1;
3. 状态转移方程
   dp[i][j] = s[i] == s[j] && dp[i+1][j-1]

判断条件整合：
a. 若: j - i == 0 , 则dp[i][j] = (s[i] == s[j]) 恒为true, 1个字符自然回文
b. 若: j - i == 1 , 则dp[i][j] = s[i] == s[j] 2个字符，首尾相等即可
c. 若: j - i == 2 , 则dp[i][j] = s[i] == s[j] 3个字符，首尾相等即可，中间有1个字符，不打紧
d. 若: j - i > 2 , 则dp[i][j] = s[i] == s[j] && dp[i + 1][j - 1] 4个字符(含)，首尾相等，外加中间子串为回文才行
a. b. c. 判断条件整合 => s[i] == s[j] && (j - i <= 2)
与d. 合并后 判断条件整合为一个 => s[i] == s[j] && ( (j - i <= 2) || dp[i + 1][j - 1] )
2、示例
input: s = "babad";
用j遍历行，i遍历列，只使用了下三角

3、代码实现

public class Solution3 {
    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0) return s;
        String res = "";
        boolean[][] dp = new boolean[s.length()][s.length()];
        int max = 0;
        for (int j = 0; j < s.length(); j++) {
            for (int i = 0; i <= j; i++) {
                dp[i][j] = s.charAt(i) == s.charAt(j) &&
                        ((j - i <= 2) || dp[i + 1][j - 1]);
                if (dp[i][j] && (j - i + 1 > max)) {
                    max = j - i + 1;
                    res = s.substring(i, j + 1);  // [i, j+1)
                }
            } // end of inner for loop
        } // end of outer for loop
        return res;
    }
}

time: O(n^2)
space: O(n^2)
4、优化使用一维数组

/*
    对Solution3的优化，使用一维数组
 */
public class Solution4 {

    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0) return s;
        String res = "";
        boolean[] dp = new boolean[s.length()];
        int n = s.length();
        for (int i = n - 1; i >= 0; i--) {
            for (int j = n - 1; j >= i; j--) {
                dp[j] = s.charAt(i) == s.charAt(j) &&
                        ((j - i) < 3 || dp[j - 1]);
                if (dp[j] && j - i + 1 > res.length())
                    res = s.substring(i, j + 1); // [i, j+1)
            }
        }
        return res;
    }
}

time: O(n^2)
space: O(n)

标签：String,int,res,牛客,length,&&,机试,HJ85,dp
来源： https://www.cnblogs.com/junstat/p/16177300.html