0
点赞
收藏
分享

微信扫一扫

LeetCode每日一题(2022/2/12)

花海书香 2022-02-12 阅读 62

1020. 飞地的数量

思路:dfs找出每个连通块的数量,如果一个连通块所有点都不在边界答案就加上这个连通块内所有点。

代码:

class Solution {
public:
    
    bool st[510][510] = {0};

    vector<int> dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1};
    void dfs (int n, int m, int x, int y, bool &f, int &cnt,vector<vector<int>>& grid) {
        
        cnt ++;
        if(x == 0 || y == 0 || x == n - 1 || y == m - 1) f = true;
        st[x][y] = true;
        for(int i = 0; i < 4; ++ i) {
            int a = x + dx[i], b = y + dy[i];
            if(a < 0 || a >= n || b < 0 || b >= m || st[a][b] || !grid[a][b]) continue;
            dfs(n, m, a, b, f, cnt, grid);
        }

    }

    int numEnclaves(vector<vector<int>>& grid) {

        
        int n = grid.size(), m = grid[0].size();
        int cnt = 0;
        for(int i = 0; i < n; ++ i)
            for(int j = 0; j < m; ++ j)
                if(grid[i][j] == 1 && !st[i][j]) {
                    bool f = false;
                    int ans = 0;
                    dfs(n, m, i, j, f, ans, grid);
                    //cout << ans << '\n';
                    if(!f) cnt += ans;

                }
        return cnt;
    }
};
举报

相关推荐

0 条评论