[LeetCode] 1721. Swapping Nodes in a Linked List

轉自LeetCode

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 <= 105
• 0 <= Node.val <= 100

Solution

可以用 19. Remove Nth Node From End of List 的思路

也是用快慢指針來找

但這次因為要 swap，所以要分兩次找

找到相對應的 node 後，swap 其值即可

kotlin

	/**
	* Example:
	* var li = ListNode(5)
	* var v = li.`val`
	* Definition for singly-linked list.
	* class ListNode(var `val`: Int) {
	* var next: ListNode? = null
	* }
	*/
	class Solution {
	fun swapNodes(head: ListNode?, k: Int): ListNode? {
	var curNode = head
	var firstNode = head
	var secondNode = head
	IntRange(1,k-1).forEach {
	curNode = curNode?.next
	}
	firstNode = curNode
	while(curNode?.next != null) {
	curNode = curNode?.next
	secondNode = secondNode?.next
	}
	//>> swap
	firstNode!!.`val` = secondNode!!.`val`.also { secondNode!!.`val` = firstNode!!.`val` }
	return head
	}
	}

view raw swapping_nodes_in_a_linked_list.kt hosted with ❤ by GitHub