[LeetCode] 957. Prison Cells After N Days_Medium Tag: Array
2021-07-29 10:54 Johnson_强生仔仔 阅读(29) 评论(0) 编辑 收藏 举报There are 8
prison cells in a row and each cell is either occupied or vacant.
Each day, whether the cell is occupied or vacant changes according to the following rules:
- If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
- Otherwise, it becomes vacant.
Note that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors.
You are given an integer array cells
where cells[i] == 1
if the ith
cell is occupied and cells[i] == 0
if the ith
cell is vacant, and you are given an integer n
.
Return the state of the prison after n
days (i.e., n
such changes described above).
Example 1:
Input: cells = [0,1,0,1,1,0,0,1], n = 7 Output: [0,0,1,1,0,0,0,0] Explanation: The following table summarizes the state of the prison on each day: Day 0: [0, 1, 0, 1, 1, 0, 0, 1] Day 1: [0, 1, 1, 0, 0, 0, 0, 0] Day 2: [0, 0, 0, 0, 1, 1, 1, 0] Day 3: [0, 1, 1, 0, 0, 1, 0, 0] Day 4: [0, 0, 0, 0, 0, 1, 0, 0] Day 5: [0, 1, 1, 1, 0, 1, 0, 0] Day 6: [0, 0, 1, 0, 1, 1, 0, 0] Day 7: [0, 0, 1, 1, 0, 0, 0, 0]
Example 2:
Input: cells = [1,0,0,1,0,0,1,0], n = 1000000000 Output: [0,0,1,1,1,1,1,0]
Constraints:
cells.length == 8
cells[i]
is either0
or1
.1 <= n <= 109
Code:
1. brute force: Time limit out T: O(n * len(cells)) S: O(len(cells))
class Solution: def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]: for _ in range(n): cells = self.nextDay(cells) return cells def nextDay(self, cells): cells_length = len(cells) newCells = [0] * cells_length for i in range(1, cells_length - 1): newCells[i] = 1 if cells[i - 1] == cells[i + 1] else 0 return newCells
2. maybe 有loop,那么可以用visited 利用空间来换时间, 还是Time limit out: O:T(n) S: O(n)
Code
class Solution: def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]: cell_length = len(cells) visited = dict() for _ in range(n): cells = self.nextDay(cells, visited) return cells def nextDay(self, cells, visited): cells_length = len(cells) if tuple(cells) not in visited: newCells = [0] * cells_length for i in range(1, cells_length - 1): newCells[i] = 1 if cells[i - 1] == cells[i + 1] else 0 visited[tuple(cells)] = newCells return visited[tuple(cells)]
3. 想法就是如果有loop,那么我们可以找到steps_loop, 然后用 n %= steps_loop, 来加快进程,如果有loop的话, 用visited = {tuple(cells) : n}, 不停 n-= 1, 一旦tuple(cells) in visited,
那么我们就知道有loop,并且steps_loop = visited[tuple(cells)] - n.
class Solution: def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]: cell_length = len(cells) visited, hasCircle = dict(), False while n > 0: if not hasCircle: state_key = tuple(cells) if state_key in visited: steps_circle= visited[state_key] - n n %= steps_circle hasCircle = True else: visited[state_key] = n if n > 0: cells = self.nextDay(cells) n -= 1 return cells def nextDay(self, cells): cells_length = len(cells) newCells = [0] * cells_length for i in range(1, cells_length - 1): newCells[i] = 1 if cells[i - 1] == cells[i + 1] else 0 return newCells