交叉字符串

给出三个字符串:s1、s2、s3，判断s3是否由s1和s2交叉构成。

您在真实的面试中是否遇到过这个题？

Yes

样例

比如 s1 = "aabcc" s2 = "dbbca"

    - 当 s3 = "aadbbcbcac"，返回  true.

    - 当 s3 = "aadbbbaccc"， 返回 false.

点题：还是边界条件

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