【leetcode】957. Prison Cells After N Days

题目如下：

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:
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]
Note:

cells.length == 8

cells[i] is in {0, 1}

1 <= N <= 10^9

解题思路：当我看到第二个用例中 N = 1000000000 后，直觉告诉我变换结果中应该存在周期性的循环。所以我就测试了前100天结果，发现这个周期是14天。这也是是一种取巧的方法了。

代码如下：

class Solution(object):def prisonAfterNDays(self, cells, N):""":type cells: List[int]:type N: int:rtype: List[int]"""if N != 0:N = N % 14 if N % 14 != 0 else 14while N > 0:tl = [0]for i in range(1,len(cells)-1):if cells[i-1] == cells[i+1]:tl.append(1)else:tl.append(0)tl.append(0)cells = tl[:]N -= 1return cells

转载于:https://www.cnblogs.com/seyjs/p/10135651.html

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