305. Number of Islands II

题目：

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?

链接： http://leetcode.com/problems/number-of-islands-ii/

题解：

又是一道Union Find的经典题。这道题代码主要参考了yavinci大神。风格还是princeton Sedgewick的那一套。这里我们可以把二维的Union-Find映射为一维的Union Find。使用Quick-Union就可以完成。但这样的话Time Complexity是O(kmn)。 想要达到O(klogmn)的话可能还需要使用Weighted-Quick Union配合path compression。二刷一定要实现。

Time Complexity - O(mn * k)， Space Complexity - O(mn)

public class Solution {
    int[][] directions = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
    
    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        List<Integer> res = new ArrayList<>();
        if(m < 0 || n < 0 || positions == null) {
            return res;
        }
        int[] id = new int[m * n];          // union find array
        int count = 0;
        Arrays.fill(id, -1);
        
        for(int i = 0; i < positions.length; i++) {
            int index = n * positions[i][0] + positions[i][1];
            if(id[index] != -1) {
                res.add(count);
                continue;
            }
            
            id[index] = index;
            count++;
            
            for(int[] direction : directions) {
                int x = positions[i][0] + direction[0];
                int y = positions[i][1] + direction[1];
                int neighborIndex = n * x + y;
                if(x < 0 || x >= m || y < 0 || y >= n || id[neighborIndex] == -1) {
                    continue;
                }
                if(!connected(id, index, neighborIndex)) {
                    union(id, neighborIndex, index);
                    count--;    
                }
            }
            
            res.add(count);
        }
        return res;
    }
    
    private boolean connected(int[] id, int p, int q) {
        return id[p] == id[q];
    }
    
    private void union(int[] id, int p, int q) {
        int pid = id[p];
        int qid = id[q];
        for(int i = 0; i < id.length; i++) {
            if(id[i] == pid) {
                id[i] = qid;
            }
        }
    }
}

 

二刷:

加入了Path compression以及Weight， 速度快了不少。

Time Complexity - (k * logmn)  Space Complexity - O(mn),  这里k是positions的长度

public class Solution {
    private int[] id;
    private int[] sz;
    private int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        List<Integer> res = new ArrayList<>();
        if (positions == null || positions.length == 0 || m < 0 || n < 0) {
            return res;
        }
        id = new int[m * n]; 
        sz = new int[m * n];
        for (int i = 0; i < id.length; i++) {
            id[i] = i;
        }
        
        int count = 0;
        for (int[] position : positions) {
            int p = position[0] * n + position[1];
            sz[p]++;
            count++;
            for (int[] direction : directions) {
                int newRow = position[0] + direction[0];
                int newCol = position[1] + direction[1];
                if (newRow < 0 || newCol < 0 || newRow > m - 1 || newCol > n - 1) {
                    continue;
                }
                int q = newRow * n + newCol;
                if (sz[q] > 0) {
                    if (isConnected(p, q)) {
                        continue;    
                    } else {
                        union(p, q);
                        count--;
                    }
                }
            }
            res.add(count);
        }
        return res;
    }
    
    private int getRoot(int p) {
        while (p != id[p]) {
            id[p] = id[id[p]];
            p = id[p];
        } 
        return p;
    }
    
    private boolean isConnected(int p, int q) {
        return getRoot(p) == getRoot(q);
    }
    
    private void union(int p, int q) {
        int rootP = getRoot(p);
        int rootQ = getRoot(q);
        if (rootP == rootQ) {
            return;
        } else {
            if (sz[p] < sz[q]) {
                id[rootP] = rootQ;
                sz[q] += sz[p];
            } else {
                id[rootQ] = rootP;
                sz[p] += sz[q];
            }
        }
    }
}

 

 

 

Reference:

https://leetcode.com/discuss/69392/python-clear-solution-unionfind-class-weighting-compression

https://www.cs.princeton.edu/~rs/AlgsDS07/01UnionFind.pdf

https://leetcode.com/discuss/69397/my-simple-union-find-solution

https://leetcode.com/discuss/69572/easiest-15ms-java-solution-written-mins-with-explanations

https://leetcode.com/discuss/69585/union-find-java-implements

https://leetcode.com/discuss/69374/solution-using-union-find-path-compression-weight-balancing

https://leetcode.com/discuss/70392/java-union-find-solution

https://leetcode.com/discuss/72435/share-my-java-union-find-solution

https://leetcode.com/discuss/69513/simple-python-not-normal-union-find

http://algs4.cs.princeton.edu/15uf/