LeetCode 305. Number of Islands II（小岛）

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本文探讨了在二维网格中动态增加陆地并计算岛屿数量的问题，利用并查集实现高效的岛屿合并操作，并通过压缩路径优化算法提高性能。

原题网址：https://leetcode.com/problems/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:

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?

方法：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。

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;
    }
}