题目
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 <= 10^5
0 <= Node.val <= 100
题目大意
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)。
解题思路
- 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。
代码
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
}