Description
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.
Note:
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
*/
#include<iostream>
#include<stdlib.h>
using namespace std;
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if(head==NULL)
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;
else
ptr_before->next = ptr_nth->next;
return head;
}
};
Note
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.