从m个小朋友报数,报到1的小朋友出圈
时间: 2024-05-06 20:19:48 浏览: 16
,剩下的小朋友重新从1开始报数,直到最后只剩下一个小朋友为止。这个问题可以使用循环链表来解决。
首先,我们创建一个循环链表,将所有小朋友的编号依次加入链表中。然后,从链表的头部开始,每次数到第k个小朋友,就将该小朋友出圈,并从链表中删除。接着,重新从当前位置开始,继续数k个小朋友,再将该小朋友出圈,直到链表中只剩下一个小朋友为止。
以下是Python代码实现:
```python
class Node:
def __init__(self, data):
self.data = data
self.next = None
class CircularLinkedList:
def __init__(self):
self.head = None
def add_node(self, data):
new_node = Node(data)
if self.head is None:
self.head = new_node
new_node.next = self.head
else:
current = self.head
while current.next != self.head:
current = current.next
current.next = new_node
new_node.next = self.head
def delete_node(self, node):
if self.head is None:
return
if self.head == node:
current = self.head
while current.next != self.head:
current = current.next
current.next = self.head.next
self.head = self.head.next
else:
current = self.head
prev = None
while current.next != self.head:
prev = current
current = current.next
if current == node:
prev.next = current.next
break
def josephus(self, k):
current = self.head
while current.next != current:
count = 1
while count != k:
current = current.next
count += 1
self.delete_node(current)
current = current.next
return current.data
if __name__ == '__main__':
n = 10 # 小朋友的数量
k = 3 # 报数到3的小朋友出圈
cll = CircularLinkedList()
for i in range(1, n + 1):
cll.add_node(i)
winner = cll.josephus(k)
print("胜利者是第", winner, "个小朋友")
```
输出:
```
胜利者是第 4 个小朋友
```