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Remove Nth Node From End of List

Tags : linkedlist, pointers, leetcode, cpp, medium

Problem Statement - link #

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

Examples #

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 2:

Input: head = [1,2], n = 1
Output: [1]

Constraints #

Follow up: Could you do this in one pass?

Solutions #

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        
        int i = 1;
        
        ListNode* left = head;
        
        while( left->next != NULL){
            left = left->next;
            i++;
        }
        
        if(n>i)return NULL;
        
        if(n==i)return head->next;
        
        int j = i - n - 1 ;
        
        left = head;
        
        while(j--)left = left -> next;

        left->next = left->next->next;

        return head;
    }
};