leetcode(java) 5. Longest Palindromic Substring

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于 2023-01-18 16:28:17 首次发布

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文章介绍了一种利用动态规划算法来找出给定字符串中的最长回文子串的方法。通过二维布尔数组记录子串是否为回文，从单字符开始逐步扩展到更长子串，最终找到最长的回文子串。提供的Java代码示例展示了具体的实现过程。

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Q: Given a string s, return the longest palindromic substring in s.

Input: s = "babad"
Output: "bab"
Explanation: "aba" is also a valid answer.

A: I have two solution.

Solution1 :

I will use dynamic programming to solve this problem.

Two demension array f[i] [j] represents whether the substring s[i...j] is palindromic.

transfer equation:

when i==j, f[i][i] = true;

when i!=j, f[i][j] = (s[i]==s[j]) && f[i+1][j-1] ? true: false;

In order to claculate all the substring， I calculate the substring from length 1 to length n.

Otherwise, it can not calculate the array correctly.

public String longestPalindrome(String s) {
        int n = s.length();
        String res = s.substring(0,1);
        if(s.length()<2){
            return s;
        }
        boolean[][] f = new boolean[n][n];
        for(int i=0;i<n;i++){
            f[i][i] = true;
        }

        for(int i=0;i<n-1;i++){
            if( s.charAt(i)==s.charAt(i+1)){
                f[i][i+1] = true;
                if(res.length()<2){
                    res = s.substring(i,i+2);
                }
            }
        }

        for(int l=2;l<n;l++){
            for(int i=0;i<n && i+l<n;i++){
                int j = i+l;
                if(s.charAt(i)==s.charAt(j)){
                    f[i][j] = f[i+1][j-1];
                    if(f[i][j] && (j-i+1)>res.length()){
                        res = s.substring(i,j+1);
                    }
                }
            }
        }
        return res;
    }

Solution 2:

I will firstly determine the number of middle element.

There are two possibilities: one or two (a, or aa)

Then I wil loop the whole String and judge whether it is a palindromic.

class Solution {
    public String longestPalindrome(String s) {
        int n = s.length();
        if(n < 2){
            return s;
        }
        String res = s.substring(0,1);
        for(int i=0;i<n;i++){
            String ans = judge(i,i,s);
            if(ans.length()>res.length()){
                res = ans;
            }
        }

        for(int i=0;i<n-1;i++){
            if(s.charAt(i)!=s.charAt(i+1)) continue;
            String ans = judge(i,i+1,s);
            if(ans.length()>res.length()){
                res = ans;
            }
        }

        return res;
    }   

    public String judge(int begin, int end, String s){
        int n = s.length();
        while(begin>=0 && end<n && s.charAt(begin)==s.charAt(end)){
            begin--;
            end++;
        }
        return s.substring(begin+1, end);
    }
}