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.

//利用dp解决，dp[i][j]的状态表示为s1[0...i] + s2[0...j]的字符串区间能否组成s3.//那么，动态转移方程为:// 1） s1[i-1] == s3[i+j-1] && dp[i-1][j] = true 那么，dp[i][j] = true;// 2） s2[j-1] == s3[i+j-1] && dp[i][j-1] = true 那么，dp[i][j] = true;class Solution {public:    bool isInterleave(std::string s1, std::string s2, std::string s3) {if(s1.size() + s2.size() != s3.size()) return false;std::vector<std::vector<bool>> dp(s1.size()+1,std::vector<bool>(s2.size()+1,0));dp[0][0] = 1;for (int i = 1; i < s1.size()+1; i++){if(s1[i-1] == s3[i-1] && dp[i-1][0]) dp[i][0] = true;}for (int i = 1; i < s2.size()+1; i++){if(s2[i-1] == s3[i-1] && dp[0][i-1]) dp[0][i] = true;}for (int i = 1; i < s1.size() + 1; i++){for (int j = 1; j < s2.size() + 1; j++){if(s1[i-1] == s3[i+j-1] && dp[i-1][j]) dp[i][j] = true;if(s2[j-1] == s3[i+j-1] && dp[i][j-1]) dp[i][j] = true;}}return dp[s1.size()][s2.size()];    }};