【LeetCode】206. 反转一个单向链表

问题描述

Reverse a singly linked list.

A linked list can be reversed either iteratively or recursively. Could you implement both?

反转一个单向链表。

链表可以迭代或递归地反转。你能实现这两个吗?

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

Python 实现

迭代实现：

# 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
        """
        # Iteratively.
        tmpNode = None
        preNode = None
        curNode = head
        while curNode != None:
            # Marked with pointer.
            preNode = ListNode(curNode.val)
            
            # Update temp list.
            preNode.next = tmpNode
            tmpNode = preNode
            
            # Continue to the next node.
            curNode = curNode.next
        
        return tmpNode

递归实现：

# 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
        """

        # Recursively.
        return self.recurse(head, None)
    
    def recurse(self, curNode, tmpNode):
        if curNode == None:
            return tmpNode
        
        # Update temp list.
        head = ListNode(curNode.val)
        head.next = tmpNode
        
        return self.recurse(curNode.next, head)
            
        

链接：https://leetcode.com/problems/reverse-linked-list/