LeetCode: Interleaving String

第一次dfs， 没过large， 马上想到dp， 一次过了

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 (s3.size() != s1.size()+s2.size()) return false;
        int m = s1.size();
        int n = s2.size();
        vector<vector<bool>> f(m+1, vector<bool>(n+1));
        f[0][0] = true;
        for (int i = 1; i <= m; i++) {
            f[i][0] = s1[i-1] == s3[i-1]? f[i-1][0]:false;
        }
        for (int i = 1; i <= n; i++) {
            f[0][i] = s2[i-1] == s3[i-1]? f[0][i-1]:false;
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; 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[m][n];
    }
};

 C#

public class Solution {
    public bool IsInterleave(string s1, string s2, string s3) {
        if (s3.Length != s1.Length + s2.Length) return false;
        bool[,] f = new bool[s1.Length+1, s2.Length+1];
        f[0, 0] = true;
        for (int i = 1; i <= s1.Length; i++) {
            f[i, 0] = s1[i-1] == s3[i-1]? f[i-1, 0] : false;
        }
        for (int i = 1; i <= s2.Length; i++) {
            f[0, i] = s2[i-1] == s3[i-1]? f[0, i-1] : false;
        }
        for (int i = 1; i <= s1.Length; i++) {
            for (int j = 1; j <= s2.Length; 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.Length, s2.Length];
    }
}