Interleaving String

https://leetcode.com/problems/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.

这题有点类似于之前的求编辑距离的题：http://www.cnblogs.com/biyeymyhjob/archive/2012/09/28/2707343.html

定义f[i][j]为 以s1[0，i]s2[0，j]匹配s3[i][j]

当j==0时，f[i][0]= f[i-1][0] && s1[i-1]==s3[i-1]

当i==0时，f[0][j]= f[0][j-1] && s2[j-1]==s3[j-1]

当i>=1&&j>=1时，f[i][j]= (f[i-1][j] && s1[i-1]==s3[i-1][j]) || (f[i][j-1] && s2[j-1]== s3[i][j-1])

bool isInterleave(string s1, string s2, string s3) {
        if(s3.length()!=s1.length()+s2.length())
           return false;

        vector<vector<bool>> f(s1.length()+,vector<bool>(s2.length()+,true));
        for(int i=;i<=s1.length();i++)
           f[i][]= f[i-][] && s1[i-]==s3[i-];

        for(int i=;i<=s2.length();i++)
           f[][i]=f[][i-] && s2[i-]==s3[i-];

        for(int i=;i<=s1.length();i++)
        {
            for(int j=;j<=s2.length();j++)
            {
                f[i][j]=(f[i-][j] && s1[i-]==s3[i+j-]) || (f[i][j-] && s2[j-]==s3[i+j-]);
            }
        }

        return f[s1.length()][s2.length()];
    }