Number of Islands II

最新推荐文章于 2020-09-26 11:09:32 发布

原创最新推荐文章于 2020-09-26 11:09:32 发布 · 545 阅读

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#need another try #union-find

本文介绍了一种算法，用于计算二维网格中岛屿的数量，并能在每次添加陆地后更新岛屿数量。通过使用并查集（Union Find）数据结构，实现高效处理。挑战在于达到O(klogmn)的时间复杂度。

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?

class UnionFind {
    vector<int> id;
    vector<int> rank;
    int count;
    
public:
    UnionFind( int n ): count(n) {
        id.resize(n, 0);
        rank.resize(n,0);
        for( int i = 0; i < n; i++ ) {
            id[i] = i;
        }
    }
    
    int find( int i ) {
        while( i != id[i]) {
            id[i] = id[id[i]];
            i = id[i];
        }
        return i;
    }
    
    bool is_connected(int p, int q) {
        int i = find(p);
        int j = find(q);
        if( i != j) return false;
        return true;
    }
    void connect ( int p , int q) {
        int i = find(p);
        int j = find(q);
        if (i == j) return;
        if (rank[i] < rank[j]) id[i] = j;
        else if (rank[i] > rank[j]) id[j] = i;
        else {
            id[i] = j;
            rank[j]++;
        }
        count--;
    }
    
};
class Solution {
public:
    vector<int> numIslands2(int m, int n, vector<pair<int, int>>& positions) {
        vector<vector<int>> matrix(m, vector<int>(n,0));
        
        vector<vector<int>> dirs = { {1,0}, {-1,0}, {0,1}, {0,-1} };
        vector<int> rst;
        UnionFind uf(m*n);
        int count = 0;
        for( auto p : positions ) {
            matrix[p.first][p.second] = 1;
            int idx = p.first*n + p.second;
            count++;
            bool flag = false;
            for( auto dir : dirs ) {
                int row = p.first + dir[0], col = p.second + dir[1];
                if(row < 0 || row >= m || col < 0 || col >= n ) continue; 
                
                if(matrix[row][col] == 1 ) {
                    int neighb = row*n + col;
                    if(!uf.is_connected(idx, neighb))count--;
                    flag = true;
                    uf.connect(idx, neighb);
                }
            } 
            rst.push_back(count);
        }
        return rst;
    }
};