leetcode 97. 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.

第一个思路肯定可能就是递归了

class Solution {
public:
    int mark = 0;
    void dfs(string& s1, string& s2, string& s3, int a, int b, int c) {
        if (c == s3.size() && a == s1.size() && b == s2.size()) {
            mark = 1;
            return;
        }
        for (int i = c; i < s3.size(); ++i) {
            if (s1[a] == s3[c]) {
                dfs(s1, s2, s3, a+1, b, c+1);
            }
            if (s2[b] == s3[c]) {
                dfs(s1, s2, s3, a, b+1, c+1);
            }
        }
    }
    bool isInterleave(string s1, string s2, string s3) {
        dfs(s1, s2, s3, 0, 0, 0);
        if (mark) return true;
        return false;
    }
    
};

但是这样会超时，时间复杂度最坏将近O(n^3) 所以采用dp的方法，能降到O(n^2)的时间复杂度。那么如何dp呢？其实很简单啦
构造一个匹配矩阵，ForExample

可以想象，向右走代表匹配s1[i]和s3[k]向下走代表匹配s2[j]和s3[k],如果能匹配上就记1，那么如果dp[n][m]为1就代表能匹配上。上图给出了一个匹配路径。
那么转移方程如下：
dp[i][j]=dp[i-1][j]&(s1[i]==s3[i+j-1])|(dp[i][j-1]&(s2[j]==s3[i+j-1])
代码如下：

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