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.)
Wedescribe 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]isin{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
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