力扣hot100 岛屿数量 dfs 图论

Problem: 200. 岛屿数量

文章目录

• 思路
• 复杂度
• Code

思路

复杂度

时间复杂度: O ( n ) O(n) O(n)

空间复杂度: O ( 1 ) O(1) O(1)

Code

class Solution {
	int n, m;

	public int numIslands(char[][] grid)
	{
		n = grid.length;
		if (n == 0)
			return 0;
		m = grid[0].length;
		int ans = 0;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				if (grid[i][j] == '1')
				{
					dfs(grid, i, j);
					ans++;
				}
		return ans;
	}

	private void dfs(char[][] g, int x, int y)
	{
		if (x < 0 || x >= n || y < 0 || y >= m || g[x][y] == '0')
			return;
		g[x][y] = '0';
		int[] dx = { 0, 1, 0, -1 };
		int[] dy = { 1, 0, -1, 0 };
		for (int i = 0; i < 4; i++)
			dfs(g, x + dx[i], y + dy[i]);
	}
}