5. Longest Palindromic Substring 题解

最新推荐文章于 2025-07-16 16:59:19 发布

原创最新推荐文章于 2025-07-16 16:59:19 发布 · 282 阅读

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本文介绍了一种利用动态规划求解最长回文子串的方法，通过避免重复计算提高效率。文中详细阐述了状态转移方程及边界条件，并给出了具体实现代码。

题目：

Total Accepted: 202411
Total Submissions: 805973
Difficulty: Medium
Contributor: LeetCode

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"

动态规划方法：
我们要知道如何避免不必要的重新计算，验证回文。如果我们已经知道的'bab”''bab”是回文，那么可知，“'ababa”''ababa”必是回文，因为左右两端字母是一样的。
定义p（i，j）p（i，j）如下：
P(i,j)={true,if the substring Si…Sj is a palindromefalse,otherwise. P(i,j)={true,if the substring Si…Sj is a palindromefalse,otherwise.
因此，
P(i, j) = ( P(i+1, j-1) \text{ and } S_i == S_j )P(i,j)=(P(i+1,j−1) and Si==Sj)
基本情况如下：
P(i, i) = trueP(i,i)=true
P(i, i+1) = ( S_i == S_{i+1} )P(i,i+1)=(Si==Si+1)
复杂度分析
.Time complexity : O(n^2)O(n2). This gives us a runtime complexity of O(n^2)O(n2).
.Space complexity : O(n^2)O(n2). It uses O(n^2)O(n2) space to store the table.

这是以abade为例，得到的dp矩阵：

给出代码：

public String
 longestPalindrome(String s) {

    boolean[][]
 flag = new boolean[s.length()][s.length()];

    int maxlen
 = 0,start
 = 0;

    for(int i
 = 0;i
 < s.length(); i++){

        flag[i][i]
 = true;

        maxlen
 = 1;

        start
 = i;

    }

    for(int i
 = 0;i
 < s.length()-1;
 i++)

        if(s.charAt(i)==s.charAt(i+1)){

            flag[i][i+1]
 = true;

            maxlen
 = 2;

            start
 = i;

        }

    for(int len
 = 3;
 len<= s.length(); len++)

        for(int i
 = 0;i
 < s.length()-len+1;
 i++){

            int j
 = i+len-1;

            if(s.charAt(i)==s.charAt(j)&&flag[i+1][j-1]==true){

                flag[i][j]
 = true;

                maxlen
 = len;

                start
 = i;

            }

        }

    return s.substring(start,
 start+maxlen);

}