【leetcode】957. Prison Cells After N Days

題目如下：

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:

1. cells.length == 8
2. cells[i] is in {0, 1}
3. 1 <= N <= 10^9

解題思路：當我看到第二個用例中 N = 1000000000 后，直覺告訴我變換結果中應該存在周期性的循環。所以我就測試了前100天結果，發現這個周期是14天。這也是是一種取巧的方法了。

代碼如下：

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 14
        while 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 -= 1
        return cells