Interleaving String

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.

------------------------------------- thinking ---------------------------------------
The below solution is beautiful!!!!!!
https://oj.leetcode.com/discuss/19973/8ms-c-solution-using-bfs-with-explanation
Please try to give some optimizations in the codes to reduce the usage of the memory

Ordinary DP solution
https://oj.leetcode.com/discuss/11694/my-dp-solution-in-c

-------------------------------------- codes ----------------------------------------------
class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int len1 = s1.size();
        int len2 = s2.size();
        int len3 = s3.size();
        if (len1+len2 != len3) {
            return false;
        }
        bool dp[len1+1][len2+1];
        memset(dp, false, sizeof(dp));
        for(int i = 0; i < len1+1; i++) {
            for (int j = 0; j < len2+1; j++) {
                if (i==0 && j==0) {
                    dp[0][0] = true;
                } else if (i==0) {
                    dp[0][j] = dp [0][j-1] && (s2[j-1] == s3[j-1]);
                } else if (j==0) {
                    dp[i][0] = dp[i-1][0] && (s1[i-1] == s3[i-1]);
                } else {
                    //bug here-> please be careful about the index, it should be s2[j-1] == s3[i+j-1]
                    //instead of s2[j] == s3[i+j]  or s2[j-1] == s3[i+j-2]
                    dp[i][j] = (dp[i][j-1] && (s2[j-1] == s3[i+j-1])) ||
                               (dp[i-1][j] && (s1[i-1] == s3[i+j-1]));
                }
            }
        }
        return dp[len1][len2];
    }
};