19. Remove Nth Node From End of List #

题目 #

Given the head of a linked list, remove the nth node from the end of the list and return its head.

Follow up: Could you do this in one pass?

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]

Constraints:

The number of nodes in the list is sz.
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

题目大意 #

删除链表中倒数第 n 个结点。

解题思路 #

这道题比较简单，先循环一次拿到链表的总长度，然后循环到要删除的结点的前一个结点开始删除操作。需要注意的一个特例是，有可能要删除头结点，要单独处理。

这道题有一种特别简单的解法。设置 2 个指针，一个指针距离前一个指针 n 个距离。同时移动 2 个指针，2 个指针都移动相同的距离。当一个指针移动到了终点，那么前一个指针就是倒数第 n 个节点了。

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
 * }
 */

// 解法一
func removeNthFromEnd(head *ListNode, n int) *ListNode {
	dummyHead := &ListNode{Next: head}
	preSlow, slow, fast := dummyHead, head, head
	for fast != nil {
		if n <= 0 {
			preSlow = slow
			slow = slow.Next
		}
		n--
		fast = fast.Next
	}
	preSlow.Next = slow.Next
	return dummyHead.Next
}

// 解法二
func removeNthFromEnd1(head *ListNode, n int) *ListNode {
	if head == nil {
		return nil
	}
	if n <= 0 {
		return head
	}
	current := head
	len := 0
	for current != nil {
		len++
		current = current.Next
	}
	if n > len {
		return head
	}
	if n == len {
		current := head
		head = head.Next
		current.Next = nil
		return head
	}
	current = head
	i := 0
	for current != nil {
		if i == len-n-1 {
			deleteNode := current.Next
			current.Next = current.Next.Next
			deleteNode.Next = nil
			break
		}
		i++
		current = current.Next
	}
	return head
}

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