[LeetCode] 200. Number of Islands

岛屿的数量。题意是给一个二维矩阵，由0和1组成，1代表岛屿。请输出岛屿的数量。例子，

Example 1:

Input:
11110
11010
11000
00000

Output: 1

Example 2:

Input:
11000
11000
00100
00011

Output: 3

题意很直观，做法也很直观，分别是DFS和BFS。都是先序遍历的思路，我就直接给代码了。唯一需要解释的地方是BFS的做法，我将坐标变成了code，这样方便记录。DFS的做法更好记。

DFS

时间O(mn)

空间O(mn) - worse case where it's all 0 in the input

 1 /**
 2  * @param {character[][]} grid
 3  * @return {number}
 4  */
 5 var numIslands = function (grid) {
 6     let res = 0;
 7     let m = grid.length;
 8     if (m === 0) return 0;
 9     let n = grid[0].length;
10 
11     var dfs = function (grid, i, j) {
12         if (i < 0 || j < 0 || i >= m || j >= n || grid[i][j] === '0') return;
13         grid[i][j] = '0';
14         dfs(grid, i, j + 1);
15         dfs(grid, i, j - 1);
16         dfs(grid, i - 1, j);
17         dfs(grid, i + 1, j);
18     }
19 
20     for (let i = 0; i < m; i++) {
21         for (let j = 0; j < n; j++) {
22             if (grid[i][j] === '1') {
23                 dfs(grid, i, j);
24                 res++;
25             }
26         }
27     }
28     return res;
29 };

 

BFS

时间O(mn)

空间O(mn)

 1 /**
 2  * @param {character[][]} grid
 3  * @return {number}
 4  */
 5 var numIslands = function (grid) {
 6     let res = 0;
 7 
 8     var bfs = function (grid, x, y) {
 9         grid[x][y] = '0';
10         let n = grid.length;
11         let m = grid[0].length;
12         let queue = [];
13         let code = x * m + y;
14         queue.push(code);
15         while (queue.length) {
16             code = queue.shift();
17             let i = parseInt(code / m);
18             let j = code % m;
19             if (i > 0 && grid[i - 1][j] == '1') {
20                 queue.push((i - 1) * m + j);
21                 grid[i - 1][j] = '0';
22             }
23             if (i < n - 1 && grid[i + 1][j] == '1') {
24                 queue.push((i + 1) * m + j);
25                 grid[i + 1][j] = '0';
26             }
27             if (j > 0 && grid[i][j - 1] == '1') {
28                 queue.push((i * m) + j - 1);
29                 grid[i][j - 1] = '0';
30             }
31             if (j < m - 1 && grid[i][j + 1] == '1') {
32                 queue.push((i * m) + j + 1);
33                 grid[i][j + 1] = '0';
34             }
35         }
36     }
37 
38     for (let i = 0; i < grid.length; i++) {
39         for (let j = 0; j < grid[0].length; j++) {
40             if (grid[i][j] === '1') {
41                 bfs(grid, i, j);
42                 res++;
43             }
44         }
45     }
46     return res;
47 };