LeadCode19. 删除链表的倒数第N个结点

LeadCode19. 删除链表的倒数第N个结点

1. 问题描述

2. 思路

1. slow和fast指针都指向head
2. fast指针先向前走n步
3. slow和fast指针同时向前走，直到fast指针指向最后一个节点（fast.Next == nil），此时slow指针指向倒数第n+1个节点
4. 删除slow.Next节点，即slow.Next = slow.Next.Next

3. 代码

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func removeNthFromEnd(head *ListNode, n int) *ListNode {
    slow, fast := head, head
    for i := 0; i < n; i++ {
        fast = fast.Next
    }

    if fast == nil {
        return head.Next
    }

    for fast.Next != nil {
        slow = slow.Next
        fast = fast.Next
    }

    slow.Next = slow.Next.Next
    return head
}