Leetcode解题: Number of Islands (200)

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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

解法1：DFS

用DFS遍历每一个元素，如果是1就表示有一个岛屿，并且用DFS遍历所有与之相连的节点。并且把他们标注为非岛屿。
Java

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public class Solution {
    public int numIslands(char[][] grid) {
   		    if (grid.length == 0 || grid[0].length == 0) {
				return 0;
			}
			int res = 0;
			for (int i = 0; i < grid.length; ++i) {
				for (int j = 0; j < grid[0].length; ++j) {
					if (grid[i][j] == '1') {
						++res;
						dfs(grid, i, j);
					}
				}
			}
			return res;
    }
	void dfs (char[][] grid, int i, int j) {
		if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length) {
			return;
		}
		if (grid[i][j] == '1') {
			grid[i][j] = '0';
			dfs(grid, i + 1, j);
			dfs(grid, i - 1, j);
			dfs(grid, i, j + 1);
			dfs(grid, i, j - 1);
		}
		return;
	}		
}

解法2：BFS

Java

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class pair {
	int x, y;
	public pair(int row, int col) {
		this.x = row;
		this.y = col;
	}
};
public class Solution {
    public int numIslands(char[][] grid) {
		if (grid.length == 0 || grid[0].length == 0) {
			return 0;
		}
		int res = 0;
		for (int i = 0; i < grid.length; ++i) {
			for (int j = 0; j < grid[0].length; ++j) {
				if (grid[i][j] == '1') {
					++res;
					bfs(grid, i, j);
				}
			}
		}    
		return res;
    }
    
    private int[][] directions = new int[][] {{0,1},{0,-1},{1,0},{-1,0}};
	private void bfs(char[][] grid, int i, int j) {
	    int row = grid.length;
	    int col = grid[0].length;
		Queue<pair> queue = new LinkedList<pair>();
		if (grid[i][j] == '1') {
			queue.offer(new pair(i, j));
			grid[i][j] = '0';
		}
		while (!queue.isEmpty()) {
			pair cur = queue.poll();
			for (int d = 0; d < directions.length; ++d) {
			    int x = cur.x + directions[d][0];
			    int y = cur.y + directions[d][1];
			    if (!isInbound(row, col, x, y)) {
			       continue;
			    }
			    if (grid[x][y] == '1') {
			        queue.offer(new pair(x,y));
			        grid[x][y] = '0';
			    }
			}
		}		
		return;
	}
	
	private boolean isInbound(int row, int col, int i, int j) {
	   return i >= 0 && i < row && j >= 0 && j < col;
	}
}