[Leetcode] 305. Number of Islands II （并查集）

最新推荐文章于 2021-11-30 06:01:53 发布

Haskei

最新推荐文章于 2021-11-30 06:01:53 发布

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本文链接：https://blog.youkuaiyun.com/Haskei/article/details/107906327

本文介绍如何使用并查集算法解决动态增加陆地后的岛屿数量计算问题，通过实例展示算法过程，包括初始化地图、执行操作及计算岛屿数量。

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305. Number of Islands II （并查集）

Hard

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
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

Follow up:

Can you do it in time complexity O(k log mn), where k is the length of the positions?

思路：

参考 https://www.cnblogs.com/grandyang/p/5190419.html

class Solution:
    def numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:
        roots = [-1] * (m * n)

        def findRoot(n):
            if n == roots[n]:
                return n
            roots[n] = findRoot(roots[n])
            return roots[n]

        cnt = 0
        rlt = []
        for px, py in positions:
            num = px * n + py
            if roots[num] != -1:
                rlt.append(cnt)
                continue
            roots[num] = num
            cnt += 1
            direction = [[0, -1], [-1, 0], [0, 1], [1, 0]]
            for a, b in direction:
                x, y = a + px, b + py
                curr_num = x * n + y
                if x >= 0 and x < m and y >= 0 and y < n and roots[curr_num] != -1:
                    p, q = findRoot(curr_num), num
                    if p != q:
                        roots[p] = q
                        cnt -= 1
            rlt.append(cnt)

        return rlt