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 = 3, positions = [[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[0][0] 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[0][1] 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[1][2] 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[2][1] 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?

Number of Islands變形，這題是一個點一個點的增加，最開始初始化時沒有陸地，每增加一個點，都要統一現在總共的島嶼個數。

使用Union-Find對小島進行合並。並查集記得要進行壓縮（islands[island] = islands[islands[island]]; ），速度會快很多。

Java:

public class Solution {  
    private int[] islands;  
    private int root(int island) {  
        while (islands[island] != island) {  
            islands[island] = islands[islands[island]];  
            island = islands[island];  
        }  
        return island;  
    }  
    private int[] yo = {-1, 1, 0, 0};  
    private int[] xo = {0, 0, -1, 1};  
    public List<Integer> numIslands2(int m, int n, int[][] positions) {  
        islands = new int[m*n];  
        Arrays.fill(islands, -1);  
        int island = 0;  
        List<Integer> nums = new ArrayList<>();  
        for(int i=0; i<positions.length; i++) {  
            int y =positions[i][0];  
            int x = positions[i][1];  
            int id=y*n+x;  
            islands[id] = id;  
            island ++;  
            for(int j=0; j<4; j++) {  
                int ny = y+yo[j];  
                int nx = x+xo[j];  
                int nid=ny*n+nx;  
                if (ny>=0 && ny<m && nx>=0 && nx<n && islands[nid] != -1) {  
                    int root = root(nid);  
                    if (root != id) {  
                        islands[root] = id;  
                        island --;  
                    }  
                }  
            }  
            nums.add(island);  
        }  
        return nums;  
    }  
}  

類似題目：

[LeetCode] 200. Number of Islands 島嶼的數量

[LeetCode] 323. Number of Connected Components in an Undirected Graph 無向圖中的連通區域的個數

[LeetCode] 305. Number of Islands II 島嶼的數量之二