LeetCode[200] Number of Islands

最新推荐文章于 2025-08-18 21:45:06 发布

原创最新推荐文章于 2025-08-18 21:45:06 发布 · 186 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #dfs

本文介绍了一种使用深度优先搜索(DFS)算法来计算二维网格中岛屿数量的方法。通过遍历每个网格元素，如果遇到陆地('1')则递增岛屿计数，并启动DFS将相连的陆地标记为已访问('0')，避免重复计数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:

Answer: 1

Example 2:

Answer: 3

每遇到一个'1'点就进入dfs，每次遍历到一个点就将其置'0'，当周围4个点没有1时dfs返回，此时算作一个island。

将矩阵每个点作为入口点遍历一遍得到island的个数

class Solution {
public:
	int numIslands(vector<vector<char>>& grid) {
		int count = 0;
		for (int i = 0; i < grid.size(); ++i) {
			for (int j = 0; j < grid[0].size(); ++j) {
				if (grid[i][j] == '1') {
					++count;
					dfs(grid, i, j);
				}
			}
		}
		return count;
	}
	void dfs(vector<vector<char>>& grid, int x, int y) {
		grid[x][y] = '0';
		if (x + 1 < grid.size() && grid[x + 1][y] == '1')
			dfs(grid, x + 1, y);
		if (x - 1 >= 0 && grid[x - 1][y] == '1')
			dfs(grid, x - 1, y);
		if (y + 1 < grid[0].size() && grid[x][y + 1] == '1')
			dfs(grid, x, y + 1);
		if (y - 1 >= 0 && grid[x][y - 1] == '1')
			dfs(grid, x, y - 1);
	}
};