1、題目：環形單鏈表的約瑟夫問題

普通思路：時間復雜度O（n × m)

代碼：

class Node:
    def __init__(self,value):
        self.value = value
        self.next = None
def test(head,m):
    if not head or head.next == head or m < 2:
        return head
    # last = head
    # while last.next != head:
    #     last = last.next
 

    count = 2
    node = head
    while node.next != node:
        if count == m:
            # last = node.next
            node.next = node.next.next
            count = 1
        node = node.next
        count += 1
    head.next = None
    # head = node
    return node
head = Node(1)
head.next = Node(2)
head.next.next 
 
= Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
head.next.next.next.next.next = Node(6)
head.next.next.next.next.next.next = Node(7)
head.next.next.next.next.next.next.next = head
m = 3
test(head,m)

遞歸思路：

從1人環的0計算到10人環，結果為4。轉化公式：

由圖知，10人環中最後入海的是4號，現由其在1人環中的對應編號0來求解。

公式：其中，m為報數值，i為第幾輪。

代碼：

class Node:
    def __init__(self,value):
        self.value = value
        self.next = None
def josephus(head,m):
    last = head
    n = 1
    while last.next != head:
        last = last.next
        n += 1
    fn = 0
    for i in range(2,n+1):
        fn = (fn + m) % i
    last = head
    if fn > 1:
        for i in range(fn-1):
            last = last.next
    pre = last.next 
    last.next = None
    pre.next = pre
    return pre
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
head.next.next.next.next.next = Node(6)
head.next.next.next.next.next.next = Node(7)
head.next.next.next.next.next.next.next = head
m = 3
josephus(head,m)

鏈表問題（4）----環形鏈表