【Leetcode】Interleaving String - 动态规划二维转一维

最新推荐文章于 2025-12-20 20:19:18 发布

转载最新推荐文章于 2025-12-20 20:19:18 发布 · 275 阅读

文章标签：

#leetcode #Java #动态规划 #String

本文通过Java实现动态规划算法解决字符串交织问题，详细解释了递归式和具体实现步骤，包括初始化、循环和判断条件，展示了如何通过比较三个字符串s1、s2和s3来判断s3是否由s1和s2交织而成。

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,
Given:
s1 = "aabcc",
s2 = "dbbca",

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

Java:

http://blog.youkuaiyun.com/linhuanmars/article/details/24683159

动态规划

递归式：

res[i][j] = res[i-1][j]&&s1.charAt(i-1)==s3.charAt(i+j-1) || res[i][j-1]&&s2.charAt(j-1)==s3.charAt(i+j-1)

public class Solution {  
    public boolean isInterleave(String s1, String s2, String s3) {  
       // Note: The Solution object is instantiated only once and is reused by each test case.  
        if (s1 == null || s2 == null || s3 == null) return false;  
        if (s1.length() + s2.length() != s3.length()) return false;  
        String minWord=s1.length()>s2.length()?s2:s1;
        String maxWord=s1.length()>s2.length()?s1:s2;
        boolean[] res=new boolean[minWord.length()+1];
        res[0]=true;
        for(int i=0;i<minWord.length();i++)
        {
            res[i+1]=res[i]&&minWord.charAt(i)==s3.charAt(i);
        }
        for(int i=0;i<maxWord.length();i++)
        {
            res[0]=res[0]&&maxWord.charAt(i)==s3.charAt(i);
            for(int j=0;j<minWord.length();j++)
            {
                
                res[j+1]=res[j+1]&&maxWord.charAt(i)==s3.charAt(i+j+1)||res[j]&&minWord.charAt(j)==s3.charAt(i+j+1);
            }
        }
        return res[minWord.length()]; 
        
    }  
}