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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
}