97. Interleaving String

最新推荐文章于 2021-01-30 02:13:05 发布

转载最新推荐文章于 2021-01-30 02:13:05 发布 · 54 阅读

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原文链接：http://www.cnblogs.com/renyualbert/p/5891242.html

本文介绍了一种用于判断字符串s3是否由s1和s2交错组成的算法。通过递归及动态规划两种方法实现，旨在帮助理解字符串匹配问题的解决思路。

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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.

public bool IsInterleave(string s1, string s2, string s3) {
        if(s1 == "") return s2 == s3;
        if(s2 == "") return s1 == s3;
        if(s1.Length + s2.Length != s3.Length) return false;
        return Dp(s1,s2,s3,0,0,0);
    }
    
    private bool Dp(string s1, string s2, string s3, int i1,int i2, int i3)
    {
        if(i3 == s3.Length) return true;
        else if(i1 >= s1.Length && i2 >= s2.Length) return false;
        else
        {
            bool judgeS1 = false;
            bool judgeS2 = false;
            if(i1 <s1.Length && s1[i1] == s3[i3])
            {
               judgeS1 = Dp(s1,s2,s3,i1+1,i2,i3+1);
            }
            if(judgeS1) return true;
            if(i2 < s2.Length && s2[i2] == s3[i3])
            {
                judgeS2 = Dp(s1,s2,s3,i1,i2+1,i3+1);
            }
            return  judgeS2;
        }
    }

public bool IsInterleave(string s1, string s2, string s3) {
        if(s1 == "") return s2 == s3;
        if(s2 == "") return s1 == s3;
        if(s1.Length + s2.Length != s3.Length) return false;
        var dp = new bool[s1.Length+1,s2.Length+1];
        dp[0,0] = true;
        for(int i = 1;i<= s2.Length;i++)
        {
            dp[0,i] = dp[0,i-1] && (s2[i-1] == s3[i-1]); 
        }
        for(int i = 1;i<= s1.Length;i++)
        {
            for(int j = 0;j<= s2.Length;j++)
            {
                if(j ==0) dp[i,j] = dp[i-1,j] && (s1[i-1] == s3[i-1]);
                else
                {
                    dp[i,j] = (dp[i-1,j]&&(s1[i-1] == s3[j+i-1]))|| (dp[i,j-1]&&(s2[j-1] == s3[j+i-1]));
                }
            }
        }
        return dp[s1.Length,s2.Length];
    }

转载于:https://www.cnblogs.com/renyualbert/p/5891242.html