[Leetcode] 0305. Number of Islands II

Explore diverse LeetCode solutions in Python, C++, JavaScript, SQL, and TypeScript. Ideal for interview prep, learning, and code practice in multiple programming languages. Github Repo Link

内容目录

Hard

You are given an empty 2D binary grid grid of size mxn. The grid represents a map where 0's represent water and 1's represent land. Initially, all the cells of grid are water cells (ie, 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 where positions[i] = [r _i , c _i ] is the position (r _i , c _i ) at which we should operate the i ^th operation.

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

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 1:

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

Example 2:

Input: m = 1, n = 1, positions = [[0,0]]
Output: [1]

Constraints:

1 <= m, n, positions.length <= 10 ⁴
1 <= m * n <= 10 ⁴
positions[i].length == 2
0 <= r _i < m
0 <= c _i < n

Follow up: Could you solve it in time complexity O(k log(mn)), where k == positions.length?

Python

				
					# 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))

Hard

Python

相关文章