题目链接

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

题目原文

Reverse a singly linked list.

题目翻译

翻转单向链表

思路方法

这个题目比较基础，解法非常多，能AC的解法也很多，这里只整理部分思路以供参考。

思路一

利用栈结构，将链表内容依次压入栈，再从栈依次弹出即可构造逆序。下面的代码用普通数组模拟栈。

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 """
        p = head
        newList = []
        while p:
            newList.insert(0, p.val)
            p = p.next

        p = head
        for v in newList:
            p.val = v
            p = p.next
        return head

思路二

与思路一栈的思想类似，不过是直接在原链表操作，通过迭代将节点重组，前面的节点转移到重组链表的后面。实际上就是头结点倒插操作。

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 """
        new_head = None
        while head:
            p = head
            head = head.next
            p.next = new_head
            new_head = p
        return new_head

思路三

递归，注意终止条件

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 """
        if not head or not head.next:
            return head

        p = head.next
        n = self.reverseList(p)

        head.next = None
        p.next = head
        return n

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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