链表中间节点快慢指针的基础应用

最新推荐文章于 2024-04-14 21:02:27 发布

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最新推荐文章于 2024-04-14 21:02:27 发布

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分类专栏：算法文章标签：链表 python 算法单元测试

本文链接：https://blog.youkuaiyun.com/weixin_39255587/article/details/106333163

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本文深入探讨了链表中点查找的四种不同算法实现，包括返回中点、上中点、下中点及其前一个节点的方法。通过对比验证，确保算法的正确性，并提供了随机链表生成和测试函数。

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基本方法：使用快慢指针，慢指针每次向后移动一个节点，快指针移动两个节点。针对特殊要求再修改快慢指针的起始位置等细节。对数器选择：使用列表保存每个节点，直接求中间节点，比对两种方法的输出。

1）输入链表头节点，奇数长度返回中点，偶数长度返回上中点

def midOrUpMid(head):
    if not head:
        return
    slow = fast = head
    while fast.next and fast.next.next:
        slow = slow.next
        fast = fast.next.next
    return slow

def varifyMidOrUpMid(head):
    if not head:
        return
    lst = []
    cur = head
    while cur:
        lst.append(cur)
        cur = cur.next
    return lst[(len(lst)-1)>>1]

2）输入链表头节点，奇数长度返回中点，偶数长度返回下中点

def midOrDownMid(head):
    if not head or not head.next:
        return head
    slow = fast = head.next
    while fast.next and fast.next.next:
        slow = slow.next
        fast = fast.next.next
    return slow

def varifyMidOrDownMid(head):
    if not head:
        return
    lst = []
    cur = head
    while cur:
        lst.append(cur)
        cur = cur.next
    return lst[(len(lst))>>1]

3）输入链表头节点，奇数长度返回中点前一个，偶数长度返回上中点前一个

def preMidOrUpMid(head):
    if not head or not head.next or not head.next.next:
        return
    slow = head
    fast = head.next.next
    while fast.next and fast.next.next:
        slow = slow.next
        fast = fast.next.next
    return slow

def varifyPreMidOrUpMid(head):
    if not head:
        return
    lst = []
    cur = head
    while cur:
        lst.append(cur)
        cur = cur.next
    if len(lst) > 2:
        return lst[(len(lst)-3)>>1]

4）输入链表头节点，奇数长度返回中点前一个，偶数长度返回下中点前一个

def preMidOrDownMid(head):
    if not head or not head.next:
        return
    slow = head
    fast = head.next
    while fast.next and fast.next.next:
        slow = slow.next
        fast = fast.next.next
    return slow

def varifyPreMidOrDownMid(head):
    if not head:
        return
    lst = []
    cur = head
    while cur:
        lst.append(cur)
        cur = cur.next
    if len(lst) > 1:
        return lst[(len(lst)-2)>>1]

生成随机链表

def generateLinkedList(maxnum=20):
	import random
	import string
    num = random.randint(0, maxnum)
    if num == 0:
        return None
    head = Node(random.choice(string.ascii_uppercase))
    cur = head
    for i in range(1, num):
        cur.next = Node(random.choice(string.ascii_uppercase))
        cur = cur.next
    return head

测试

def test(num=10000):
    for i in range(num):
        head = generateLinkedList(30)
        if midOrUpMid(head) != varifyMidOrUpMid(head):
            print("varifyMidOrUpMid Failed!")
            traverseLinkedList(head)
            break
        if midOrDownMid(head) != varifyMidOrDownMid(head):
            print("varifyMidOrDownMid Failed!")
            traverseLinkedList(head)
            break
        if preMidOrUpMid(head) != varifyPreMidOrUpMid(head):
            print("varifyPreMidOrUpMid Failed!")
            traverseLinkedList(head)
            break
        if preMidOrDownMid(head) != varifyPreMidOrDownMid(head):
            print("varifyPreMidOrDownMid Failed!")
            traverseLinkedList(head)
            break
    else:
        print("nice!")
        
if __name__ == "__main__":
    test()