LeetCode 957. Prison Cells After N Days

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Words count in article: 345 | Reading time ≈ 2

题目

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.)

We describe the current state of the prison in the following way: cells[i] == 1 if the i-th cell is occupied, else cells[i] == 0.

Given the initial state of the prison, return the state of the prison after N days (and N such changes described above.)

Example 1:

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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:

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Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]

Note:

  • cells.length == 8
  • cells[i] is in {0, 1}
  • 1 <= N <= 10^9

思路

Period.

代码

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class Solution(object):
    def prisonAfterNDays(self, cells, N):
        """
        :type cells: List[int]
        :type N: int
        :rtype: List[int]
        """
        count = 0
        N %= 14
        if N == 0:
            N = 14
        while count < N:
            newCell = [0] * 8
            for i in range(1, 7):
                if cells[i - 1] == cells[i + 1]:
                    newCell[i] = 1
                else:
                    newCell[i] = 0
            cells = newCell
            count += 1
        return cells