题目

restore-ip-addresses/

算法

* DP

O（n^2）

代码

class Solution {

public:

    bool isInterleave(string s1, string s2, string s3) {

        if (s1.size() + s2.size() != s3.size()) return false;

        int n1 = s1.size();

        int n2 = s2.size();

        vector<vector<bool> > dp(n1 + 1, vector<bool> (n2 + 1, false)); 

        dp[0][0] = true;

        for (int i = 1; i <= n1; ++i) {

            dp[i][0] = dp[i - 1][0] && (s1[i - 1] == s3[i - 1]);

        }

        for (int i = 1; i <= n2; ++i) {

            dp[0][i] = dp[0][i - 1] && (s2[i - 1] == s3[i - 1]);

        }

        for (int i = 1; i <= n1; ++i) {

            for (int j = 1; j <= n2; ++j) {

                dp[i][j] = (dp[i - 1][j] && s1[i - 1] == s3[i - 1 + j]) || (dp[i][j - 1] && s2[j - 1] == s3[j - 1 + i]);

            }

        }

        return dp[n1][n2];

    }

};