Leetcode 97. Interleaving String

题目描述：

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.

bool isInterleave(string s1, string s2, string s3) {        
        if(s3.size() != s1.size() + s2.size()){
            return false;
        }
        return isInterleave(begin(s1),end(s1),begin(s2),end(s2),begin(s3),end(s3));
    }
template
bool isInterleave(Iter first1,Iter last1,Iter first2,Iter last2,Iter first3,Iter last3){
    if(first3 == last3){
        return first1 == last1 && first2 == last2;
    }
   return (*first1 == *first3 && isInterleave(next(first1),last1,first2,last2,next(first3),last3 ))  || ( *first2 == *first3 && isInterleave(first1,last1,next(first2),last2,next(first3),last3));
}

最直观的解法就是这样，但是会超时。

bool isInterleave(string s1, string s2, string s3) {
        vector> f(s1.size() + 1,vector(s2.size() + 1,true));
        if(s1.size() + s2.size() != s3.size()){
            return false;
        }
        for(int i = 1;i <= s1.size();i++){
            f[i][0] = f[i - 1][0] && s1[i - 1] == s3[i - 1];
        }
        for(int j = 1;j <= s2.size();j++){
            f[0][j] = f[0][j - 1] && s2[j - 1] == s3[j - 1];
        }
        for(int i = 1;i <= s1.size();i++){
            for(int j = 1;j <= s2.size();j++){
                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]);
            }
        }
        return f[s1.size()][s2.size()];
    }