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// @Title: 反转链表 (Reverse Linked List)
// @Author: 15816537946@163.com
// @Date: 2019-11-20 22:12:10
// @Runtime: 0 ms
// @Memory: 2.5 MB
/*
* @lc app=leetcode.cn id=206 lang=golang
*
* [206] 反转链表
*
* https://leetcode-cn.com/problems/reverse-linked-list/description/
*
* algorithms
* Easy (63.65%)
* Likes: 541
* Dislikes: 0
* Total Accepted: 86.9K
* Total Submissions: 136.1K
* Testcase Example: '[1,2,3,4,5]'
*
* 反转一个单链表。
*
* 示例:
*
* 输入: 1->2->3->4->5->NULL
* 输出: 5->4->3->2->1->NULL
*
* 进阶:
* 你可以迭代或递归地反转链表。你能否用两种方法解决这道题?
*
*/
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
// 原地重建
/*
func reverseList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
p := reverseList(head.Next)
head.Next.Next = head
head.Next = nil
return p
}
func reverseList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
prev := new(ListNode)
for head != nil {
next := head.Next // 先保存下一个节点的状态
head.Next = prev
prev = head
head = next
}
return prev
}
*/
// 迭代解法, 原地重建链表
func reverseList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
// 记录逆序后的状态
// pre := new(ListNode)
var pre *ListNode = nil
for head != nil {
next := head.Next
head.Next = pre
pre = head
head = next
}
return pre
}
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