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.)
解题思路
先找到状态的循环，直接找到最后的状态
代码
class Solution:
def prisonAfterNDays(self, cells: List[int], N: int) -> List[int]:
dic = {}
while N > 0:
dic[''.join(map(str, cells))] = N
N -= 1
tmp = [0] * 8
for i in range(1, 7):
tmp[i] = 1 if cells[i - 1] == cells[i + 1] else 0
cells = tmp
t = ''.join(map(str, cells))
if t in dic:
N = N % (dic[t] - N)
return cells
class Solution:
def prisonAfterNDays(self, cells: List[int], N: int) -> List[int]:
def find(cells):
tmp = [0] * 8
for i in range(1, 7):
tmp[i] = 1 if cells[i - 1] == cells[i + 1] else 0
return tmp
cycle = []
state = find(cells)
while state not in cycle:
cycle.append(state)
state = find(state)
return cycle[(N - 1) % len(cycle)]
