【牛客题霸题解】删除链表中的倒数第n个结点

这道题目其实和输出链表的倒数第n个结点是基本上一致的。https://blog.nowcoder.net/n/4e57664df6974486bc352aa5a1e4613c
和上一道题一样，定义快慢指针，快指针先走n步，然后快慢指针一起走，快指针走到空的时候慢指针正好在倒数第n个结点上。
区别的地方只是在于要删除某个结点的话，我们需要知道它的前一个结点是什么。
如果要删除的结点是链表头结点的话（这个时候没有前一个结点）。这种情况下我们进行一下特判就可以了。
c++

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        // write code here
        ListNode* fast = head;
        ListNode* slow = head;
        ListNode* pre = head;
        for(int i = 0 ; i < n ; i++)
        {
            fast = fast->next;
        }
        while(fast!=NULL)
        {
            fast = fast->next;
            pre = slow;
            slow = slow->next;
        }
        ListNode* target;
        if(slow!=head)
        {
            target = pre->next;
            pre->next = target->next;
        }
        else{
            target = head;
            head = head->next;
        }
        delete target;
        return head;
    }
};

java

import java.util.*;

public class Solution {

    public ListNode removeNthFromEnd (ListNode head, int n) {
        ListNode fast = head;
        ListNode slow = head;
        ListNode pre = head;
        for(int i = 0 ; i < n ; i++){
            fast = fast.next;
        }
        while(fast!=null){
            fast = fast.next;
            pre = slow;
            slow = slow.next;
        }
        ListNode target;
        if(slow!=head){
            target = pre.next;
            pre.next = target.next;
        }
        else{
            target = head;
            head = head.next;
        }
        return head;
    }
}

python

class Solution:
    def removeNthFromEnd(self , head , n ):
        fast = head
        slow = head
        pre = head
        for i in range(0,n):
            fast = fast.next
        while fast!=None:
            fast = fast.next
            pre = slow;
            slow = slow.next
        if slow!=head:
            target = pre.next
            pre.next = target.next
        else:
            target = head
            head = head.next
        return head