[リートコード] 0305. 島の数 II

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難しい

You are given an empty 2D binary grid グリッド of size xn. The grid represents a map where 0‘s represent water and 1‘s represent land. Initially, all the cells of グリッド are water cells (i.e., all the cells are 0‘s).

We may perform an add land operation which turns the water at position into a land. You are given an array positions どこ positions[i] = [r_i, c_i] is the position (r _i , c _i ) at which we should operate the ^私は operation.

戻る an array of integers 答え どこ 答え[i] is the number of islands after turning the cell (r _i , c _i ) into a land.

アン島は水に囲まれており、隣接する土地を水平または垂直に接続することによって形成されます。グリッドの 4 つの端すべてが水に囲まれていると考えることができます。

例 1:

入力： m = 3, n = 3, positions = [[0,0],[0,1],[1,2],[2,1]]
出力： [1,1,2,3]
説明：
Initially, the 2d grid is filled with water.
- Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land. We have 1 island.
- Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land. We still have 1 island.
- Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land. We have 2 islands.
- Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land. We have 3 islands.

例 2:

入力： m = 1, n = 1, positions = [[0,0]]
出力： [1]

制約:

1 <= m, n, positions.length <= 10⁴
1 <= m * n <= 10⁴
positions[i].length == 2
0 <= r_i < m
0 <= c_i < n

フォローアップ： Could you solve it in time complexity O(k log(mn)), where k == positions.length?

パイソン

				
					# time complexity: O(m*n)
# space complexity: O(m*n)
from typing import List


class Solution:
    class UnionFind:
        def __init__(self, size):
            self.rep = [i for i in range(size)]
            self.rank = [0] * size
            self.count = 0

        def find(self, node):
            if node != self.rep[node]:
                self.rep[node] = self.find(self.rep[node])
            return self.rep[node]

        def union(self, node1, node2):
            rep1 = self.find(node1)
            rep2 = self.find(node2)
            if rep1 != rep2:
                if self.rank[rep1] > self.rank[rep2]:
                    self.rep[rep2] = rep1
                elif self.rank[rep2] < self.rank[rep1]:
                    self.rep[rep1] = rep2
                else:
                    self.rep[rep2] = rep1
                    self.rank[rep1] += 1
                self.count -= 1

    def numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:
        uf = self.UnionFind(m * n)
        res = []
        land_map = set()

        for row, col in positions:
            if (row, col) in land_map:
                res.append(uf.count)
                continue

            land_map.add((row, col))
            uf.count += 1

            for i, j in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
                r, c = row + i, col + j
                if r < 0 or c < 0 or r >= m or c >= n:
                    continue
                if (r, c) in land_map:
                    uf.union(row * n + col, r * n + c)

            res.append(uf.count)

        return res


m = 3
n = 3
positions = [[0, 0], [0, 1], [1, 2], [2, 1]]
print(Solution().numIslands2(m, n, positions))

難しい

パイソン

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