LeetCode – 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]

 

 

public List<Integer> numIslands2(int m, int n, int[][] positions) {
    int[] rootArray = new int[m*n];
    Arrays.fill(rootArray,-1);
 
    ArrayList<Integer> result = new ArrayList<Integer>();
 
    int[][] directions = {{-1,0},{0,1},{1,0},{0,-1}};
    int count=0;
 
    for(int k=0; k<positions.length; k++){
        count++;
 
        int[] p = positions[k];
        int index = p[0]*n+p[1];
        rootArray[index]=index;//set root to be itself for each node
 
        for(int r=0;r<4;r++){
            int i=p[0]+directions[r][0];
            int j=p[1]+directions[r][1];
 
            if(i>=0&&j>=0&&i<m&&j<n&&rootArray[i*n+j]!=-1){
                //get neighbor's root
                int thisRoot = getRoot(rootArray, i*n+j);
                if(thisRoot!=index){
                    rootArray[thisRoot]=index;//set previous root's root
                    count--;
                }
            }
        }
 
        result.add(count);
    }
 
    return result;
}
 
public int getRoot(int[] arr, int i){
    while(i!=arr[i]){
        i=arr[i];
    }
    return i;
}

 二刷：

public class Solution {
    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        int[] id = new int[m * n]; // 表示各个index对应的root
        
        List<Integer> res = new ArrayList<>();
        Arrays.fill(id, -1); // 初始化root为-1，用来标记water, 非-1表示land
        int count = 0; // 记录island的数量
        
        int[][] dirs = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
        for (int i = 0; i < positions.length; i++) {
            count++;
            int index = positions[i][0] * n + positions[i][1];           
            id[index] = index; // root初始化
            
            for (int j = 0; j < dirs.length; j++) {
                int x = positions[i][0] + dirs[j][0];
                int y = positions[i][1] + dirs[j][1];
                if (x >= 0 && x < m && y >= 0 && y < n && id[x * n + y] != -1) {
                    int root = root(id, x * n + y);

                    // 发现root不等的情况下，才union, 同时减小count
                    if (root != index) {
                        id[root] = index;
                        count--;
                    }
                }
            }
            res.add(count);
        }
        return res;
    }
    
    public int root(int[] id, int i) {
        while (i != id[i]) {
            id[i] = id[id[i]]; // 优化，为了减小树的高度                
            i = id[i];
        }
        return i;
    }
}