约瑟夫问题

class Child(object):
    first = None

    def __init__(self, no=None, pNext=None):
        self.no = no
        self.pNext = pNext

    def addChild(self):
        cur = None
        for i in range(4):
            child = Child(i + 1)  # 实例化新的孩子

            if i == 0:  # 如果是
                self.first = child
                child.pNext = self.first
                cur = self.first
            else:
                cur.pNext = child
                child.pNext = self.first
                cur = cur.pNext

    def showChild(self):
        cur = self.first
        while cur.pNext != self.first:
            print("当前孩子的编号是:%s" % cur.no)
            cur = cur.pNext
        print("当前孩子的编号是:%s" % cur.no)

    def countChild(self, m, k):
        tail = self.first
        # 当tail退出循环的时候，tail已经跑到指向first的节点
        while tail.pNext != self.first:
            tail = tail.pNext
        # 解决指定数数的孩子的起始位置的问题
        for i in range(k-1):
            tail = tail.pNext
            self.first = self.first.pNext

        while tail != self.first:
            # 数两下（迈一步），tail指向first后一个节点，first指向其前一个节点
            for i in range(m-1):
                tail = tail.pNext
                self.first = self.first.pNext
            # 因为取出数中的节点，所以要将断裂的链表连接起来
            self.first = self.first.pNext
            tail.pNext = self.first
        print("最终身下的孩子的编号是：%s" % self.first.no)


cc = Child()
cc.addChild()
cc.showChild()
cc.countChild(4, 1)