Number of Islands II

Hard

A 2d grid map ofmrows andncolumns is initially filled with water. We may perform anaddLandoperation 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 gridgridis 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 thepositions?

Analysis

思路:

使用并查集,每个格子作为一个节点。当一个格子变成岛屿,和它的四个邻居依次连接,相当于在图中加四条边

其他方法似乎都会TLE,并查集似乎是唯一能过OJ的方法。

Solution

Union Find

Use rank to do path compression

class Solution {
    class UnionFind {
        int count; // # of connected components
        int[] parent;
        int[] rank;

        public UnionFind(char[][] grid) { // for problem 200
            count = 0;
            int m = grid.length;
            int n = grid[0].length;
            parent = new int[m * n];
            rank = new int[m * n];
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (grid[i][j] == '1') {
                        parent[i * n + j] = i * n + j;
                        ++count;
                    }
                    rank[i * n + j] = 0;
                }
            }
        }

        public UnionFind(int N) { // for problem 305 and others
            count = 0;
            parent = new int[N];
            rank = new int[N];
            for (int i = 0; i < N; ++i) {
                parent[i] = -1;
                rank[i] = 0;
            }
        }

        public boolean isValid(int i) { // for problem 305
            return parent[i] >= 0;
        }

        public void setParent(int i) {
            parent[i] = i;
            ++count;
        }

        public int find(int i) { // path compression
            if (parent[i] != i) parent[i] = find(parent[i]);
            return parent[i];
        }

        public void union(int x, int y) { // union with rank
            int rootx = find(x);
            int rooty = find(y);
            if (rootx != rooty) {
                if (rank[rootx] > rank[rooty]) {
                    parent[rooty] = rootx;
                } else if (rank[rootx] < rank[rooty]) {
                    parent[rootx] = rooty;
                } else {
                    parent[rooty] = rootx; rank[rootx] += 1;
                }
                --count;
            }
        }

        public int getCount() {
            return count;
        }
    }

    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        List<Integer> ans = new ArrayList<>();
        UnionFind uf = new UnionFind(m * n);

        int[] dr = {0, 1, 0, -1};
        int[] dc = {1, 0, -1, 0};

        for (int[] pos : positions) {
            int r = pos[0], c = pos[1];
            int id = r * n + c;
            uf.setParent(id);

            for (int k = 0; k < 4; k++) {
                int nr =  r + dr[k];
                int nc =  c + dc[k];
                int nid = nr * n + nc;

                if (nr >= 0 && nr < m && nc >= 0 && nc < n) {
                    if (uf.isValid(nid)) {
                        uf.union(nid, id);
                    }
                }
            }
            ans.add(uf.getCount());
        }
        return ans;
    }
}

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