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[lintcode] problem 434 – number of islands ii

admin 11月 13, 2020 0

Given a n,m which means the row and column of the 2D matrix and an array of pair A(size k). Originally, the 2D matrix is all 0 which means there is only sea in the matrix. The list pair has k operator and each operator has two integer A[i].x, A[i].y means that you can change the grid matrix[A[i].x][A[i].y] from sea to island. Return how many island are there in the matrix after each operator.

Note

0 is represented as the sea, 1 is represented as the island. If two 1 is adjacent, we consider them in the same island. We only consider up/down/left/right adjacent.

Example

No.1

Input: n = 4, m = 5, A = [[1,1],[0,1],[3,3],[3,4]]

Output: [1,1,2,2]

Explanation:

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0.  00000
    00000
    00000
    00000
1.  00000
    01000
    00000
    00000
2.  01000
    01000
    00000
    00000
3.  01000
    01000
    00000
    00010
4.  01000
    01000
    00000
    00011

No.2

Input: n = 3, m = 3, A = [[0,0],[0,1],[2,2],[2,1]]

Output: [1,1,2,2]

Code

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public class  {
    int x;
    int y;
    Point() { x = 0; y = 0; }
    Point(int a, int b) { x = a; y = b; }
}
 
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public class UnionFind{
    private int[] id;
    private int[] size;
    private int count;

    public UnionFind(int n) {
        this.id = new int[n+1];
        this.size = new int[n+1];
        this.count = n;

        for (int i = 1; i <= n; i++) {
            id[i] = i;
            size[i] = 1;
        }
    }

    public int count() {
        return count;
    }

    public boolean union(int p, int q) {
        int pRoot = find(p);
        int qRoot = find(q);

        if (pRoot == qRoot)
            return false;

        if (size[pRoot] < size[qRoot]) {
            id[pRoot] = qRoot;
            size[qRoot] += size[pRoot];
        }
        else {
            id[qRoot] = pRoot;
            size[pRoot] += size[qRoot];
        }

        count--;
        return true;
    }

    public int find(int p) {
        while (p != id[p])
            p = id[p];

        return p;
    }
}

public List<Integer> numIslands2(int n, int m, Point[] operators) {
    List<Integer> result = new ArrayList<>();
    
    if (operators == null || operators.length == 0)
        return result;
        
    int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    Set<Integer> island = new HashSet<>();
    UnionFind uf = new UnionFind(n * m);

    for (Point operator : operators) {
        int idx = operator.x * m + operator.y;
        island.add(idx);

        for (int[] dir : dirs) {
            int x = operator.x + dir[0];
            int y = operator.y + dir[1];
            int newIdx = x * m + y;

            if (x < 0 || y < 0 || x >= n || y >= m || !island.contains(newIdx))
                continue;
            
            uf.union(idx, newIdx);
        }

        result.add(uf.count() - n * m + island.size());
    }

    return result;
}
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