LeetCode Remove Nth Node From End Of List


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

For 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.

Given n will always be valid.
Try to do this in one pass.

The original problem is here.

The original code is here.

My Solution

I solve this problem in C++, as below:

*Remove Nth Node from End of List 
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
using namespace std;

 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
class Solution {
    ListNode* removeNthFromEnd(ListNode* head, int n) {
            return NULL;
        if(head->next == NULL && n == 1)
            return NULL;
        ListNode *ptr, *ptr_nth, *ptr_before;
        int count  = 0;
        ptr = head;
        ptr_nth    = head;
        ptr_before = head;
        while(ptr->next != NULL){
            if(count > n){
                ptr_before = ptr_nth;
                ptr_nth = ptr_nth->next;
            count ++;
            ptr = ptr->next;
        if(ptr_before == ptr_nth)
            head = head->next;
            ptr_before->next = ptr_nth->next;
        return head;


To solve the problem, we need three pointers, one pointer traverse the list and the second one point the element which always keeps n distance to the first pointer. The third pointer will always before the second one. Finally, after traverse the list, remove the element which pointed by the second pointer.