Description

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?

Solution

Union-Find, time O(mn + k), space O(mn)

注意时间复杂度是O(mn + k)，因为初始化parent[]需要O(mn)的时间！

class Solution {
    public static final int[][] DIRECTIONS = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        List<Integer> res = new ArrayList<>();
        if (m < 1 || n < 1 || positions == null) {
            return res;
        }
        
        int islands = 0;
        int[] parent = new int[m * n];
        Arrays.fill(parent, -1);    // -1 represents water
        
        for (int[] pos : positions) {
            int i = pos[0];
            int j = pos[1];
            int k = i * n + j;
            parent[k] = k;  // mark k as an island
            ++islands;
            
            for (int[] d : DIRECTIONS) {
                int x = i + d[0];
                int y = j + d[1];
                int z = x * n + y;
                // make sure (x, y) is valid and is island includes k itself only
                if (x < 0 || x >= m || y < 0 || y >= n || parent[z] == -1) {
                    continue;
                }
                // find
                int kset = find(parent, k);
                int zset = find(parent, z);
                if (kset != zset) {
                    // union
                    parent[kset] = zset;
                    --islands;
                }
            }
            
            res.add(islands);
        }
        
        return res;
    }
    
    private int find(int[] parent, int x) {
        if (parent[x] == x) {
            return x;
        }
        
        parent[x] = find(parent, parent[x]);
        return parent[x];
    }
}