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.

分析：动态规划， f[i][j]表示s3[0:i+j-1]是s1[0:i-1]和s2[0:j-1]的interleaving string。递推公式如下：

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].

代码如下：

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

本题目的递推版本很容易理解，但超时。