-
Notifications
You must be signed in to change notification settings - Fork 1
/
Copy path19.删除链表的倒数第n个节点.go
62 lines (61 loc) · 1.13 KB
/
19.删除链表的倒数第n个节点.go
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
/*
* @lc app=leetcode.cn id=19 lang=golang
*
* [19] 删除链表的倒数第N个节点
*
* https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/description/
*
* algorithms
* Medium (33.50%)
* Likes: 390
* Dislikes: 0
* Total Accepted: 44.4K
* Total Submissions: 132.5K
* Testcase Example: '[1,2,3,4,5]\n2'
*
* 给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。
*
* 示例:
*
* 给定一个链表: 1->2->3->4->5, 和 n = 2.
*
* 当删除了倒数第二个节点后,链表变为 1->2->3->5.
*
*
* 说明:
*
* 给定的 n 保证是有效的。
*
* 进阶:
*
* 你能尝试使用一趟扫描实现吗?
*
*/
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNthFromEnd(head *ListNode, n int) *ListNode {
slow, fast := head, head
for n > 0 {
fast = fast.Next
n--
}
for {
if fast == nil || fast.Next == nil {
break
} else {
fast = fast.Next
slow = slow.Next
}
}
if fast == nil {
return slow.Next
} else {
slow.Next = slow.Next.Next
return head
}
}