LeetCode 题解(288): Number of Islands II

题目：

A 2d grid map of m rows and n columns is initially filled with water.We may perform anaddLand 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?

题解：

解释放在了comments里。

public class Solution {
    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        List<Integer> results = new ArrayList<Integer>();
        if(m <= 0 || n <= 0 || positions.length == 0)
            return results;
        //假设相邻的1都有相同的标识符，不相邻的1有不同的标识符，此为标识符到坐标列表的映射
        HashMap<Integer, List<Integer>> map1 = new HashMap<Integer, List<Integer>>();
        //用于快速查找某一位置的标识符
        HashMap<Integer, Integer> map2 = new HashMap<Integer, Integer>();
        //标识符
        int count = 1;
        for(int i = 0; i < positions.length; i++) {
            int r = positions[i][0], c = positions[i][1];
            int p = r * n + c;
            //收集邻居的标识符，加入到set中
            Set<Integer> candidate = new HashSet<Integer>();
            int top = r - 1 >= 0 ? (r-1) * n + c : -1;
            int bot = r + 1 < m ? (r+1) * n + c : -1;
            int left = c - 1 >= 0 ? r * n + (c - 1) : -1;
            int right = c + 1 < n ? r * n + (c + 1) : -1;
            if(map2.containsKey(top))
                candidate.add(map2.get(top));
            if(map2.containsKey(bot))
                candidate.add(map2.get(bot));
            if(map2.containsKey(left))
                candidate.add(map2.get(left));
            if(map2.containsKey(right))
                candidate.add(map2.get(right));
            //如set为空，说明没有相邻的1，加入新的1
            if(candidate.isEmpty()) {
                List<Integer> l = new ArrayList<Integer>();
                l.add(p);
                map1.put(count, l);
                map2.put(p, count);
                count++;
            //否则，需要merge邻居，统一他们的标识符
            } else {
                Iterator iter = candidate.iterator();
                int cur = (Integer)iter.next();
                while(iter.hasNext()) {
                    int old = (Integer)iter.next();
                    for(int q : map1.get(old)) {
                        map2.put(q, cur);
                        map1.get(cur).add(q);
                    }
                    map1.remove(old);
                }
                map1.get(cur).add(p);
                map2.put(p, cur);
            }
            //每次的岛数，为map1的size
            results.add(map1.size());
        }
        return results;
    }
}