Number of Islands

最新推荐文章于 2025-05-07 17:27:01 发布

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最新推荐文章于 2025-05-07 17:27:01 发布

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分类专栏：算法文章标签： 2d LeetCode

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本文介绍了一种利用深度优先搜索（DFS）算法统计二维地图中岛屿数量的方法。该算法适用于由'1'（陆地）和'0'（水域）组成的网格地图，通过遍历每个网格并标记已访问过的陆地来计数彼此不相连的陆地块，即岛屿的数量。

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Given a 2d grid map of ‘1’s (land) and ‘0’s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000
Answer: 1

Example 2:

11000
11000
00100
00011
Answer: 3

对于一个静态的地图，统计岛屿个数可以使用dfs（类似于寻找一个图中的连通块个数），算法复杂度是O(m*n)。

class Solution {
    int m, n;
    int dr[4] = {1, 0, -1, 0};
    int dc[4] = {0, -1, 0, 1};
public:    
    void dfs(int x, int y, vector<vector<char> >& grid) {
        if (x < 0 || x >= m || y < 0 || y >= n || grid[x][y] != '1') return;
        grid[x][y] = '2';
        for (int i=0; i<4; i++) {
            dfs(x+dr[i], y+dc[i], grid);
        }
    }

    int numIslands(vector<vector<char> >& grid) {
        int cnt = 0;
        m = grid.size(); 
        if (m == 0) return 0;
        n = grid[0].size();
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if (grid[i][j] == '1') {
                    dfs(i, j, grid);
                    cnt++;
                }
            }
        }
        return cnt;
    }
};

Source: LeetCode 200. Number of Islands