[LeetCode] Interleaving String

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.

 

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 Dynamic Programming String

 

 

思路一：dfs，递归，超时

Time Limit Exceeded

class Solution {
    public:

        bool dfs(int idx1, int idx2,int idx3, string s1, string s2, string s3) 
        {   
            //cout << "idx1\t" << idx1 << endl;
            //cout << "idx2\t" << idx2 << endl;
            //cout << "idx3\t" << idx3 << endl;
            if(idx3 == s3.size())
            {   
                if( idx2 == s2.size() && idx1 == s1.size())
                    return true;
                else
                    return false;
            }   
            if(s1[idx1] == s3[idx3] && dfs(idx1+1,idx2,idx3+1,s1,s2,s3))
                return true;
            if(s2[idx2] == s3[idx3] && dfs(idx1,idx2+1,idx3+1,s1,s2,s3))
                return true;
            return false;
        }   
    
        bool isInterleave(string s1, string s2, string s3) 
        {   
            if((s1.size() + s2.size()) != s3.size())
                return false;
            return dfs(0,0,0,s1,s2,s3);
        }   
};

 

思路二：dp

设状态 f[i][j]，表示 s1[0,i-1] 和 s2[0,j-1]，匹配 s3[0, i+j-1]。如果 s1 的最后一个字符等
于 s3 的最后一个字符，则 f[i][j]=f[i-1][j]；如果 s2 的最后一个字符等于 s3 的最后一个字符，
则 f[i][j]=f[i][j-1]。因此状态转移方程如下：
f[i][j] = (s1[i - 1] == s3 [i + j - 1] && f[i - 1][j])
|| (s2[j - 1] == s3 [i + j - 1] && f[i][j - 1]);

class Solution {
    public:
    
        bool isInterleave(string s1, string s2, string s3) 
        {   
            if((s1.size() + s2.size()) != s3.size())
                return false;
            //f[i][j] indicates s1[0 ~ i-1] and s2[0 ~ j-1] can consititue s3[0 ~ i+j-1]
            vector<bool> tmp(s2.size() + 1 ,false);//colum size
            vector<vector<bool> >  f(s1.size() + 1, tmp);//row size
    
            f[0][0] = true;// indicate null str + null str can constitue null str

            for(int i = 1; i <=s2.size(); i++ )
            {   
                if(f[0][i-1] && s2[i-1] == s3[i-1])
                    f[0][i] = true;
            }   

            for(int i = 1; i <=s1.size(); i++ )
            {   
                if(f[i-1][0] && s1[i-1] == s3[i-1])
                    f[i][0] = true;
            }   

            for(int i = 1; i <= s1.size(); i++)
            {   
                for(int j = 1; j <= s2.size(); j++)
                {   
                    if((f[i-1][j] && s1[i-1] == s3[i+j-1]) ||
                            (f[i][j-1] && s2[j-1] == s3[i+j-1])) 
                            f[i][j] = true;
                }   
            }   
            return f[s1.size()][s2.size()];

        }
};