Number of Islands II Leetcode

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?

果然做完island1之后这道题一遍就过了。。。动态的适合用union find

public class Solution {
    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        List<Integer> res = new ArrayList<>();
        if (positions == null || positions.length == 0 || positions[0].length == 0) {
            return res;
        }
        UF uf = new UF(m * n);
        int[][] island = new int[m][n];
        int[][] dic = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        for (int[] cor : positions) {
            int row = cor[0];
            int col = cor[1];
            island[row][col] = 1;
            uf.count += 1;
            for (int[] sur : dic) {
                int r = row + sur[0];
                int c = col + sur[1];
                if (r >= 0 && r < m && c >= 0 && c < n && island[r][c] == 1) {
                    uf.connect(row * n + col, r * n + c);
                }
            }
            res.add(uf.query());
        }
        return res;
        
    }
    class UF {
        int[] id;
        int count = 0;
        public UF(int size) {
            id = new int[size];
            for (int i = 0; i < size; i++) {
                id[i] = i;
            }
        }
        public int find(int p) {
            if (id[p] == p) {
                return p;
            }
            id[p] = find(id[p]);
            return id[p];
        }
        public void connect(int a, int b) {
            int roota = find(a);
            int rootb = find(b);
            if (roota != rootb) {
                id[roota] = rootb;
                count--;
            }
        }
        public int query() {
            return count;
        }
    }
}

top solution和我的思想差不多，只是还是不明白这个时间复杂度为啥是log。。。以后回顾。。。