[LeetCode]97. Interleaving String

[LeetCode]97. Interleaving String

题目描述

思路

动态规划
以给出的样例为例

s1 = “aabcc”
s2 = “dbbca”
s3 = “aadbbcbcac”

dp	0	1 a	2 a	3 b	4 c	5 c
0	1	1	1	0	0	0
1 d	0	0	1	1	0	0
2 b	0	0	1	1	1	0
3 b	0	0	1	0	1	1
4 c	0	0	1	0	1	0
5 a	0	0	1	0	1	1

dp数组的状态与他的前面的状态有关，前面的状态表示为上述表格中当前位置的左方或者上方，当前一位置为1时，表示到当前位置是满足的，可以生成当前的字符串，在下一位置时，就需要比较s3的对应位置是否和s1或者s2的对应位置相等

代码

#include <iostream>
#include <vector>
#include <string>

using namespace std;

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

int main() {
    Solution s;
    cout << s.isInterleave("aabcc", "dbbca", "aadbbbaccc") << endl;

    system("pause");
    return 0;
}