LeetCode 206 [Reverse Linked List]

原题

翻转一个链表

样例
给出一个链表1->2->3->null，这个翻转后的链表为3->2->1->null

解题思路

• 基础链表问题，同向双指针
• temp = head.next //让temp记住head.next指向的节点
• head.next = prev //剪短head与原来head.next之间的关系，将head.next指向prev
• prev指向head，即向前移动一步
• head指向temp，也是向前移动一步
• 方法二 Recursion

完整代码

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

# 方法一 ：Non-recursion
class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        prev = None
        while head:
            temp = head.next
            head.next = prev
            prev = head
            head = temp
        return prev

# 方法二 ：Recursion
class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None or head.next is None:
            return head
        next = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return next