LeetCode #305 - Number of Islands II

最新推荐文章于 2020-05-15 10:19:01 发布

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本文介绍了一种使用并查集数据结构解决动态岛屿数量计算问题的方法。在一个初始充满水的二维网格上，通过一系列操作将水域转换为陆地，并在每次操作后计算岛屿的数量。文章详细解释了算法原理，包括如何初始化并查集、执行查找和合并操作，以及如何更新岛屿数量。

题目描述：

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:

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

Output: [1,1,2,3]

Explanation:

Initially, the 2d grid grid is filled with water. (Assume 0 represents water and 1 represents land).

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 the positions?

由于可以在中途增加点，可能会导致两个小岛连接成一个岛，所以最好的方法是并查集。

class Solution {
public:
    vector<int> numIslands2(int m, int n, vector<pair<int, int>>& positions) {
        int count=0;
        vector<int> result;
        unordered_map<int,int> parent;
        vector<pair<int,int>> dirs={{1,0},{-1,0},{0,1},{0,-1}};
        vector<vector<int>> grid(m,vector<int>(n,0));
        for(auto position:positions)
        {
            int i=position.first, j=position.second;
            grid[i][j]=1;
            int id=i*n+j;//坐标(i,j)表示在grid中是第i*n(列数)+j个点
            parent[id]=id;//初始化根等于自己
            count++; //加入一个1时，就可以先认为加入了一个岛
            for(auto dir:dirs)
            {
                int x=i+dir.first, y=j+dir.second;
                if(x<0||x>=m||y<0||y>=n) continue;
                if(grid[x][y]==0) continue;
                int cur_id=x*n+y;
                if(find(parent,id)!=find(parent,cur_id))
                { //一旦发现当前加入的1和周围1的根不同，说明加入当前1导致两个不连通的岛合并了，所以count减一
                    count--;
                    uni(parent,id,cur_id);
                }
            }
            result.push_back(count);
        }
        return result;
    }
    
    int find(unordered_map<int,int>& parent, int x)
    {
        if(parent[x]==x) return x;
        else return find(parent,parent[x]);
    }
    
    void uni(unordered_map<int,int>& parent, int a, int b)
    {
        int x=find(parent,a);
        int y=find(parent,b);
        parent[x]=y;
    }
};