【LeetCode】面试题24. 反转链表

题目:

思路:

1、双指针。一个指向反转链表的头部，一个指向原链表的头部，每次从原链表头部摘下一个节点加入反转链表头部。两个指针同时移动

2、双指针。一个一直指向head也就是反转链表的尾部，head.next指向原链表头部，一个指向反转链表头部并移动

3、递归。head的反转可以转换成head.next子问题的反转

代码:

Python

# 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
        """
        # # 双指针移动
        # cur = None
        # pre = head
        # while pre:
        #     tmp = pre.next
        #     pre.next = cur
        #     cur = pre
        #     pre = tmp
        # return cur

        # # head作为最后一个元素指针不移动
        # if head is None:
        #     return None
        # cur = head
        # while head.next:
        #     tmp = head.next.next
        #     head.next.next = cur
        #     cur = head.next
        #     head.next = tmp
        # return cur

        # 递归(子问题, 子链表反转)
        if head is None or head.next is None:
            return head
        tmp = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return tmp

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