website/content/ChapterFour/1700~1799/1721.Swapping-Nodes-in-a-Linked-List.md
You are given the head of a linked list, and an integer k.
Return the head of the linked list after swapping the values of the kth node from the beginning and the kth node from the end (the list is 1-indexed).
Example 1:
Input: head = [1,2,3,4,5], k = 2
Output: [1,4,3,2,5]
Example 2:
Input: head = [7,9,6,6,7,8,3,0,9,5], k = 5
Output: [7,9,6,6,8,7,3,0,9,5]
Example 3:
Input: head = [1], k = 1
Output: [1]
Example 4:
Input: head = [1,2], k = 1
Output: [2,1]
Example 5:
Input: head = [1,2,3], k = 2
Output: [1,2,3]
Constraints:
n.1 <= k <= n <= 10^50 <= Node.val <= 100给你链表的头节点 head 和一个整数 k 。交换 链表正数第 k 个节点和倒数第 k 个节点的值后,返回链表的头节点(链表 从 1 开始索引)。
package leetcode
import (
"github.com/halfrost/leetcode-go/structures"
)
// ListNode define
type ListNode = structures.ListNode
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func swapNodes(head *ListNode, k int) *ListNode {
count := 1
var a, b *ListNode
for node := head; node != nil; node = node.Next {
if count == k {
a = node
}
count++
}
length := count
count = 1
for node := head; node != nil; node = node.Next {
if count == length-k {
b = node
}
count++
}
a.Val, b.Val = b.Val, a.Val
return head
}