# LeetCode 305. Number of Islands II

A 2d grid map of m rows and n columns is initially filled with water. We may perform an addLand operation which turns the water at position (row, col) into a land. Given a list of positions to operate, count the number of islands after each addLand operation. 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:

Given m = 3, n = 3positions = [[0,0], [0,1], [1,2], [2,1]].
Initially, the 2d grid grid is filled with water. (Assume 0 represents water and 1 represents land).

0 0 0
0 0 0
0 0 0


Operation #1: addLand(0, 0) turns the water at grid into a land.

1 0 0
0 0 0   Number of islands = 1
0 0 0


Operation #2: addLand(0, 1) turns the water at grid into a land.

1 1 0
0 0 0   Number of islands = 1
0 0 0


Operation #3: addLand(1, 2) turns the water at grid into a land.

1 1 0
0 0 1   Number of islands = 2
0 0 0


Operation #4: addLand(2, 1) turns the water at grid into a land.

1 1 0
0 0 1   Number of islands = 3
0 1 0


We return the result as an array: [1, 1, 2, 3]

Challenge:

Can you do it in time complexity O(k log mn), where k is the length of the positions?

Union-Find, 一个operation 添加进UnionFind2D型的islands. 看四个方向上islands里parent 对应的值，若是大于0, 就find是不是在一个根下，若不在，就union起来. 四个direction走完，把islands的当前count加入res中.

Note: there could be duplicate in the positions.

Time Complexity: O(k*logmn). k = edges.length. Find: O(logmn).

Space: O(mn).

AC Java:

 1 class Solution {
2     int [] parent;
3     int [] size;
4     int count;
5
6     public List<Integer> numIslands2(int m, int n, int[][] positions) {
7         List<Integer> res = new ArrayList<>();
8         if(positions == null || positions.length == 0){
9             return res;
10         }
11
12         parent = new int[m * n + 1];
13         size = new int[m * n + 1];
14         count = 0;
15
16         int [][] dirs = new int[][]{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
17
18         for(int [] p : positions){
19             int ind1 = p * n + p + 1;
20             if(parent[ind1] != 0){
21                 res.add(count);
22                 continue;
23             }
24
25             parent[ind1] = ind1;
26             size[ind1] = 1;
27             count++;
28
29             for(int [] dir : dirs){
30                 int x = p + dir;
31                 int y = p + dir;
32                 if(x < 0 || x >= m || y < 0 || y >= n){
33                     continue;
34                 }
35
36                 int ind2 = x * n + y + 1;
37                 if(parent[ind2] > 0 && !find(ind1, ind2)){
38                     union(ind1, ind2);
39                 }
40             }
41
42             res.add(count);
43         }
44
45         return res;
46     }
47
48     private boolean find(int i, int j){
49         return root(i) == root(j);
50     }
51
52     private int root(int i){
53         while(i != parent[i]){
54             parent[i] = parent[parent[i]];
55             i = parent[i];
56         }
57
58         return parent[i];
59     }
60
61     private void union(int i, int j){
62         int p = root(i);
63         int q = root(j);
64         if(size[p] > size[q]){
65             size[p] += size[q];
66             parent[q] = p;
67         }else{
68             size[q] += size[p];
69             parent[p] = q;
70         }
71
72         count--;
73     }
74 }
posted @ 2016-02-21 09:05  Dylan_Java_NYC  阅读(511)  评论(0编辑  收藏