Leetcode 97. Interleaving String-优快云博客

本文介绍了一个基于动态规划的算法，用于判断一个字符串s3是否可以通过交错另一个两个字符串s1和s2得到。提供了完整的Java代码实现，并通过示例说明了如何使用该算法。

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题目来源：leetcode

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.

思路：DP http://blog.youkuaiyun.com/u011095253/article/details/9248073

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s3==null||s1==null||s2==null)
            return false;
        if((s1.length()+s2.length())!=s3.length())
            return false;
        boolean dp[][]=new boolean[s1.length()+1][s2.length()+1];
        dp[0][0]=true;
        for (int i=1;i<=s2.length();i++){
            if (dp[0][i-1]==true&&s2.charAt(i-1)==s3.charAt(i-1))
            dp[0][i]=true;
        }
        for (int i=1;i<=s1.length();i++){
            if (dp[i-1][0]==true&&s1.charAt(i-1)==s3.charAt(i-1))
                dp[i][0]=true;
        }
        for (int i=1;i<=s1.length();i++)
            for (int j=1;j<=s2.length();j++){
            if ((dp[i-1][j]==true&&s1.charAt(i-1)==s3.charAt(i+j-1))||(dp[i][j-1]==true&&s2.charAt(j-1)==s3.charAt(i+j-1)))
                dp[i][j]=true;
            }
            return dp[s1.length()][s2.length()];


    }
}