0019.remove-nth-node-from-end-of-list.md

0019.Remove-Nth-Node-From-End-of-List

Description

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

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

Tags: Linked List, Two Pointers

题意

题意是让你删除链表中的倒数第 n 个数

题解

思路1

我的解法是利用双指针，这两个指针相差 n 个元素，当后面的指针扫到链表末尾的时候，自然它前面的那个指针所指向的下一个元素就是要删除的元素，即 pre.next = pre.next.next;，但是如果一开始后面的指针指向的为空，此时代表的意思就是要删除第一个元素，即 head = head.next;。

func removeNthFromEnd(head *ListNode, n int) *ListNode {
    fastNode := head
    slowNode := head
    for n > 0 {
        fastNode = fastNode.Next
        n--
    }

    if fastNode != nil {
        for fastNode.Next != nil {
            fastNode = fastNode.Next
            slowNode = slowNode.Next
        }
        slowNode.Next = slowNode.Next.Next

    } else {
        head = head.Next
    }
    return head
}

思路2

思路2 ```go

```

结语

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