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:

Input: m = 3, n = 3, positions = [[0,0], [0,1], [1,2], [2,1]]
Output: [1,1,2,3]
Explanation:

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
Follow up:

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

Solution

本题也是一道Union Find的题目,但毕竟是一道hard题目,有以下几点需要注意:

  1. 因为Union Find需要用到parent数组,需要将二维矩阵转换为一维,进行下标转换即可。
  2. 维护一个count记录当前连通分量数,每加入一个点则count+1,每合并两个连通分量则count-1

Code

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int count = 0;
public List<Integer> numIslands2(int m, int n, int[][] positions) {
if(positions == null || positions.length == 0) return new ArrayList<>();
int[][] matrix = new int[m][n];
List<Integer> res = new ArrayList<>();
int[][] direc = {{0,1}, {0, -1}, {-1, 0}, {1, 0}};
int[] parent = new int[m*n]; int[] rank = new int[m*n];
for(int i = 0; i < parent.length; i++) parent[i] = i;
for(int i = 0; i < positions.length; i++){
int x = positions[i][0]; int y = positions[i][1];
matrix[x][y] = 1;
count++;
for(int j = 0; j < direc.length; j++){
int neighbor_x = x + direc[j][0];
int neighbor_y = y + direc[j][1];
if(neighbor_x < 0 || neighbor_x >= m
|| neighbor_y < 0 || neighbor_y >=n)
continue;
if(matrix[neighbor_x][neighbor_y] == 1){
union(parent, rank, x*n + y, neighbor_x*n + neighbor_y);
}
}
res.add(count);
}
return res;
}

public void union(int[] parent, int[] rank, int x, int y){
int xr = find(parent, x); int yr = find(parent, y);
if(xr != yr){
if(rank[xr] == rank[yr]){
parent[yr] = xr;
rank[xr]++;
}else if(rank[xr] > rank[yr]){
parent[yr] = xr;
}else{
parent[xr] = yr;
}
--count;
}

}

public int find(int[] parent, int x){
if(parent[x] != x){
parent[x] = find(parent, parent[x]);
}
return parent[x];
}