岛屿的数量

16 Oct 2024 in Algorithm Pageviews

leetcode!

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例1

输入

grid = [

["1", "1", "1", "1", "0"]

["1", "1", "0", "1", "0"]

["1", "1", "0", "0", "0"]

["0", "0", "0", "0", "0"]

]

输出

1

示例2

输入

grid = [

["1", "1", "0", "0", "0"]

["1", "1", "0", "0", "0"]

["0", "0", "1", "0", "0"]

["0", "0", "0", "1", "1"]

]

输出

3

范围

• n == grid.size()
• m = grid[0].size()
• 1 <= m, n <= 300
• grid[i][j] 的值为 0 或 1

代码输出如下

class Solution {
public:
    int NumIsLands(vector<vector<char>> &grid)
    {
        int n = grid.size();
        int m = grid[0].size();
        int result = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] != '1') {
                    continue;
                }
                
                result++;
                dfs(grid, i, j);
            }
        }
        result;
    }

    vector<int> dir = {-1, 0, 1, 0, -1};
    void dfs(vector<vector<char>> &grid, int i, int j)
    {
        int n = grid.size();
        int m = grid[0].size();
        queue<pair<int, int>> que;
        que.push({i, j});
        while (!que.empty()) {
            auto [r, c] = que.front();
            que.pop();
            for (int i = 0; i < 4; i++) {
                auto x_ = r + dir[i];
                auto y_ = c + dir[i + 1];
                if (x_ < 0 || x_ >= n || y_ < 0 || y_ >= m || gird[x_][y_] == '0') {
                    continue;
                }
                
                grid[x_][y_] = '0';
                que.push({x_, y_});
            }
        }
    }
};