200. Number of Islands

最新推荐文章于 2024-01-21 17:08:30 发布

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最新推荐文章于 2024-01-21 17:08:30 发布

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分类专栏： # 算法实践

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本文介绍了一种使用深度优先搜索（DFS）算法来解决二维矩阵中岛屿计数问题的方法。通过遍历矩阵并标记已访问过的陆地，有效地统计出独立岛屿的数量。

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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:

Answer: 1

Example 2:

Answer: 3

题意：统计矩阵中岛屿的个数。

思路：dfs。

class Solution {
public:
	int numIslands(vector<vector<char>>& grid) {
		const int m = grid.size();
		if (m == 0)
			return 0;
		const int n = grid[0].size();
		int res = 0;
		for (int i = 0; i < m; i++){
			for (int j = 0; j < n; j++){
				if (grid[i][j] == '1'){
					res++;
					dfs(grid, i, j, res+1);
				}
			}
		}
		return res;
	}
private:
	void dfs(vector<vector<char>>& grid, int i, int j, int num){
		if (i >= 0 && i < grid.size() && j >= 0 && j < grid[0].size()){
			if (grid[i][j] == '1'){
				grid[i][j] = ('0' + num);
				dfs(grid, i - 1, j, num);
			    dfs(grid, i + 1, j, num);
			    dfs(grid, i, j - 1, num);
			    dfs(grid, i, j + 1, num);
			}
			
		}
	}
};

二刷：

用标志数组的思路实现。因为如果岛屿的数量大于255，上述算法不太适应。但是，上述算法不需要开辟额外空间，值得借鉴。15.62%

class Solution {
public:
	int numIslands(vector<vector<char>>& grid) {
		if (grid.empty())
			return 0;
		const int LINE = grid.size();
		const int COLUMN = grid[0].size();
		vector<vector<bool>> visited(LINE, vector<bool>(COLUMN, false));
		int res = 0;;
		for (int i = 0; i < LINE; i++){
			for (int j = 0; j < COLUMN; j++){
				if (grid[i][j] == '1'  &&  !visited[i][j]){
					dfs(grid, visited, i, j);
					res++;
				}
			}
		}
		return res;
	}
private:
	void dfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int i, int j){
		if (i<0 || i>=grid.size())
			return;
		if (j < 0 || j >= grid[0].size())
			return;

		if (grid[i][j] == '1'  &&  !visited[i][j]){
			visited[i][j] = true;
			dfs(grid, visited, i + 1, j);
			dfs(grid, visited, i, j + 1);
			dfs(grid, visited, i - 1, j);
			dfs(grid, visited, i, j - 1);
		}
	}
};