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?
200. Number of Islands 变形，这题是一个点一个点的增加，最开始初始化时没有陆地，每增加一个点，都要统一现在总共的岛屿个数。
使用Union-Find对小岛进行合并。并查集记得要进行压缩（islands[island] = islands[islands[island]]; ），速度会快很多。
Disjoint-set data structure(union-find data structure)
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;
}
}
类似题目：
[LeetCode] 200. Number of Islands 岛屿的数量
[LeetCode] 323. Number of Connected Components in an Undirected Graph 无向图中的连通区域的个数
[LeetCode] 261. Graph Valid Tree 图是否是树