LeetCode-Interleaving String

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;        }        //二维动态规划, f[i][j]表示s3的前i + j个字符是否可由s1的前i个字符和s2的前j个字符组成        //则以下任一种情况成立，f[i][j]为真,否则为假        //1.s3的前i+j-1个字符可由s1的前i-1个字符以及s2的前j个字符组成，且s3[i+j-1]==s1[i-1]        //2.s3的前i+j-1个字符可由s1的前i个字符以及s2的前j-1个字符组成，且s3[i+j-1]==s2[j-1]        bool f[1000][1000];        f[0][0] = true;        for (size_t i = 1; i <= s1.size(); ++i)        {            f[i][0] = f[i - 1][0] && s1[i - 1] == s3[i - 1];        }        for (size_t j = 1; j <= s2.size(); ++j)        {            f[0][j] = f[0][j - 1] && s2[j - 1] == s3[j - 1];        }        for (size_t i = 1; i <= s1.size(); ++i)        {            for (size_t j = 1; j <= s2.size(); ++j)            {                f[i][j] = f[i - 1][j] && s1[i - 1] == s3[i + j - 1]                    || f[i][j - 1] && s2[j - 1] == s3[i + j - 1];            }        }        return f[s1.size()][s2.size()];    }};