200. Number of Islands

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

Similar: 
  • 130. Surrounded Regions
  • 305. Number of Islands II
  • 323. Number of Connected Components in an Undirected Graph

Solution 1. DFS

 1 public class Solution {
 2     public int numIslands(char[][] grid) {
 3         if (grid.length == 0 || grid[0].length == 0)
 4             return 0;
 5             
 6         int num = 0;
 7         for (int i = 0; i < grid.length; i++) {
 8             for (int j = 0; j < grid[0].length; j++) {
 9                 if (grid[i][j] == '1') {
10                     num++;
11                     dfs(grid, i, j);
12                 }
13             }
14         }
15         return num;
16     }
17     public void dfs(char[][] grid, int i, int j) {
18         grid[i][j] = '0';
19         if (i > 0 && grid[i - 1][j] == '1') {
20             dfs(grid, i - 1, j);
21         }
22         if (j > 0 && grid[i][j - 1] == '1') {
23             dfs(grid, i, j - 1);
24         }
25         if (i < grid.length - 1 && grid[i + 1][j] == '1') {
26             dfs(grid, i + 1, j);
27         }
28         if (j < grid[0].length - 1 && grid[i][j + 1] == '1') {
29             dfs(grid, i, j + 1);
30         }
31     }
32 }

 Solution 2. Union Find

 1 /* Solution 2. Union Find
 2  * Tip: convert 2d grid to 1d
 3  */
 4 public class Solution {
 5     public int numIslands(char[][] grid) {
 6         if (grid.length == 0 || grid[0].length == 0)
 7             return 0;
 8             
 9         int m = grid.length;
10         int n = grid[0].length;
11         int cnt = 0;
12         
13         // construct points and edges
14         int[] pt = new int[m*n];
15         ArrayList<int[]> edges = new ArrayList<>();
16         
17         for (int i=0; i<m; i++) {
18             for (int j=0; j<n; j++) {
19                 int id = i * n + j;
20                 pt[id] = id;
21                 if (grid[i][j] == '1') {
22                     cnt++;
23                     if (i+1<m && grid[i+1][j] == '1') {
24                         edges.add(new int[] {id, id+n}); // down connect
25                     }
26                     if (j+1<n && grid[i][j+1] == '1') {
27                         edges.add(new int[] {id, id+1}); // right connect
28                     }
29                 }
30             }
31         }
32         
33         for (int[] edge : edges) {
34             int root1 = find(pt, edge[0]);
35             int root2 = find(pt, edge[1]);
36             
37             if (root1 != root2) {
38                 pt[root2] = root1;
39                 cnt--;
40             }
41         }
42         return cnt;
43     }
44     
45     private int find(int[] pt, int id) {
46         while (pt[id] != id) {
47             pt[id] = pt[pt[id]];
48             id = pt[id];
49         }
50         return id;
51     }
52 }
posted @ 2016-03-15 13:11  Joyce-Lee  阅读(180)  评论(0)  编辑  收藏  举报