957. Prison Cells After N Days N 天后的牢房

作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.cn/

---

@TOC

题目地址：https://leetcode.com/problems/prison-cells-after-n-days/description/

题目描述

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

题目大意

有一个数组，每次操作：如果某个位置i的左边和右边的元素相等，那么当前位置改成1；否则就是0.求N次操作之后的结果是多少。

解题方法

周期是14

写了一个多小时的题目，后来才发现周期是14.

发现周期的过程就是尝试了一下，不同的数组的循环周期是多少。试了几个之后发现是14，那就是14了。

如果知道是14之后那就好办了，先把N mod 14，然后注意了！如果mod完之后等于0，应该把N设置为14！！最后我是有时间提交通过的，但是这个地方没有想明白，所以比赛结束之后才通过的。

底下的转移方程就很简单了，直接转移。因为最多操作14次，所以很容易就过了。

代码如下：

class Solution(object):
    def prisonAfterNDays(self, oldcells, N):
        """
        :type cells: List[int]
        :type N: int
        :rtype: List[int]
        """
        cells = copy.deepcopy(oldcells)
        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

日期

2018 年 12 月 16 日 —— 周赛好难