LeetCode----Reverse Linked List

最新推荐文章于 2024-05-11 15:38:10 发布

原创最新推荐文章于 2024-05-11 15:38:10 发布 · 590 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#链表 #leetcode #链表反转

Python 同时被 3 个专栏收录

127 篇文章

订阅专栏

算法之美

95 篇文章

订阅专栏

编程之美

93 篇文章

订阅专栏

这篇博客详细解析了如何反转单链表，包括三种不同的方法：使用自身作为哑节点交换、通过两个变量pre和cur交换，以及递归方法通过函数参数存储前一个节点。

Reverse Linked List

Reverse a singly linked list.

代码：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        h = head
        p = h
        q = h.next
        isFirst = True
        
        if not head or not head.next:  # 当head为空链表或者链表只有一个值
            return head
        
        if not q:  # 当链表只有两个值时，反转头和尾指针
            q.next = h
            h.next = None
            return q
        
        while q:
            r = q.next
            if isFirst:
                isFirst = False
                p.next = None
            q.next = p
            p = q
            q = r
        return p

---------------------------------------------------------------------------------------------------------------------------

2016-05-01补充：

代码1：

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head

        pre = head
        mid = head.next
        post = mid.next
        pre.next = None
        while post:
            mid.next = pre
            pre = mid
            mid = post
            post = post.next
        mid.next = pre
        return mid

代码2：# 代码2的简洁性和易理解性优于代码1

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None:
            return head
        pre = None
        mid = head
        post = None
        while mid:
            post = mid.next
            mid.next = pre
            pre = mid
            mid = post
        return pre

递归代码：

class Solution2(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None or head.next is None:
            return head
        newhead = self.reverseList(head.next)  # 得到表尾结点为新的链表头，直接返回该节点
        # 递归回退过程中进行当前节点的反向
        tail = head.next
        tail.next = head
        head.next = None
        return newhead

Discuss中看到的好的代码：

Method 1: uses self as a dummy node for swapping

def reverseList(self, head):
    pre, pre.next, tail = self, head, head
    while tail and tail.next:
        pre.next, pre.next.next, tail.next = \
                  tail.next, pre.next, tail.next.next
    return self.next

Method 2: swapping using two variable, pre and cur

def reverseList(self, head):
    pre, cur = None, head
    while cur:
        cur.next, pre, cur = pre, cur, cur.next
    return pre

Method 3: Recursive method by storing previous node in the function argument.

def reverseList(self, head, pre = None):
    if not head: return pre
    cur, head.next = head.next, pre
    return self.reverseList(cur, head)