剑指offer 圆圈中最后剩下的数字 python实现

每年六一儿童节,牛客都会准备一些小礼物去看望孤儿院的小朋友,今年亦是如此。HF作为牛客的资深元老,自然也准备了一些小游戏。其中,有个游戏是这样的:首先,让小朋友们围成一个大圈。然后,他随机指定一个数m,让编号为0的小朋友开始报数。每次喊到m-1的那个小朋友要出列唱首歌,然后可以在礼品箱中任意的挑选礼物,并且不再回到圈中,从他的下一个小朋友开始,继续0...m-1报数....这样下去....直到剩下最后一个小朋友,可以不用表演,并且拿到牛客名贵的“名侦探柯南”典藏版(名额有限哦!!^_^)。请你试着想下,哪个小朋友会得到这份礼品呢？(注：小朋友的编号是从0到n-1)

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def LastRemaining_Solution(self, n, m):
        if n==0 or m==0:
            return -1
        pHead = self.build_circle(n)
        step = 0
        while pHead:
            step += 1
            if step == m:
                
                if pHead.next != pHead:
                    pHead.val = pHead.next.val
                    pHead.next = pHead.next.next
                else:
                    print(pHead.val)
                    return
                step = 0
            else:
                pHead = pHead.next
        return
            
    def build_circle(self, n):
        tmp = None
        pHead = None
        pRear = None
        for i in range(n):
            if tmp == None:
                tmp = ListNode(i)
                pHead = tmp
            else:
                tmp.next = ListNode(i)
                tmp = tmp.next
        tmp.next = pHead
        return pHead


if __name__ == "__main__":
    double_num = input()
    n = int(double_num[0])
    m = int(double_num[1])
    so = Solution()
    so.LastRemaining_Solution(n, m)