[Leetcode]Interleaving String

思路：刚开始的思路是从s3中一个个把字符拿出来，与s1和s2中的字符比较，如果匹配则去掉那个匹配的字符，但是问题是如果s1和s2的字符与s3中的都匹配呢？首先先到用递归，当都匹配时，递归的选择s1或者s2。但是这个递归有效率问题，有大量的重叠子问题，不用说，动态规划解决之。

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(s1.size() + s2.size() != s3.size()) return false;        int n = s1.size(), m = s2.size();        vector<vector<bool> > f(n+1, vector<bool>(m+1, false));        f[0][0] = true;        for(int i = 1; i <= n; ++i)            f[i][0] = f[i-1][0] && (s1[i-1] == s3[i-1]);         for(int i = 1; i <= m; ++i)            f[0][i] = f[0][i-1] && (s2[i-1] == s3[i-1]);         for(int i = 1; i <= n; ++i)            for(int j = 1; j <= m; ++j)            {                f[i][j] = f[i][j] || f[i-1][j] && (s1[i-1] == s3[i+j-1]);                f[i][j] = f[i][j] || f[i][j-1] && (s2[j-1] == s3[i+j-1]);            }        return f[n][m];    }};