Leetcode 957. Prison Cells After N Days

文章作者:Tyan
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1. Description

Prison Cells After N Days

2. Solution

解析:Version 1,根据变换规则可知,第一位和最后一位总是0,因此只有中间6位数在变,最大可能的变换周期为2^6。因此只要记录变换周期,因此周期中的所有状态就可得出变换结果,使用字典stat来判断每次变换是否与之前的重复,列表state记录状态变化,当出现重复状态时,计算变换的周期peroid,以及一个周期的状态变化,如果没出现周期,则直接返回变换后的结果,如果出现了,则返回计算后的状态。

  • Version 1
class Solution:
    def prisonAfterNDays(self, cells: List[int], n: int) -> List[int]:
        stat = {}
        state = []
        temp = ''.join(list(map(str, cells)))
        stat[temp] = 0
        count = 0
        pre = cells[:]
        state.append(pre)
        for i in range(n):
            count += 1
            cells[0] = 0
            cells[7] = 0
            for j in range(1, 7):
                if (pre[j-1] == 1 and pre[j+1] == 1) or (pre[j-1] == 0 and pre[j+1] == 0):
                    cells[j] = 1
                else:
                    cells[j] = 0
            temp = ''.join(list(map(str, cells)))
            if temp not in stat:
                stat[temp] = count
                pre = cells[:]
                state.append(pre)
            else:
                peroid = count - stat[temp]
                state = state[stat[temp]:]
                break
        if count == n:
            return cells
        return state[(n - count) % peroid]

Reference

  1. https://leetcode.com/problems/prison-cells-after-n-days/

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