LeetCode-19.删除链表的倒数第 N 个结点

链表#

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点

示例 1：

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输入：head = [1,2,3,4,5], n = 2
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输出：[1,2,3,5]

示例 2：

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输入：head = [1], n = 1
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输出：[]

示例 3：

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输入：head = [1,2], n = 1
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输出：[1]

自己的解法

手撕，一遍过，哈哈哈！不妄我失败了这么多次链表，虽然比较简单就是了

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/**
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 * Definition for singly-linked list.
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 * type ListNode struct {
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 *     Val int
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 *     Next *ListNode
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 * }
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 */
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func removeNthFromEnd(head *ListNode, n int) *ListNode {
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    dummy := &ListNode{Next: head}
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    cur := dummy.Next
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    count := 1
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    for cur != nil {
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        cur = cur.Next
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        count++
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    }
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    need := count - n - 1
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    cur = dummy
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    for i := 0; i < need; i++ {
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        cur = cur.Next
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    }
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    if cur.Next != nil {
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        cur.Next = cur.Next.Next
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    }
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    return dummy.Next
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}

双指针

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func removeNthFromEnd(head *ListNode, n int) *ListNode {
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  dummyNode := &ListNode{0, head}
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  fast, slow := dummyNode, dummyNode
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  for i := 0; i <= n; i++ {
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    fast = fast.Next
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  }
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  for fast != nil {
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    fast = fast.Next
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    slow = slow.Next
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  }
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  slow.Next = slow.Next.Next
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  return dummyNode.Next
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}