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:
- The number of nodes in the list is
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）。
- 这道题虽然是 medium，但是实际非常简单。题目要求链表中 2 个节点的值，无非是先找到这 2 个节点，然后再交换即可。链表查询节点需要 O(n)，2 次循环找到对应的 2 个节点，交换值即可。
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
}