Problem description:

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

Example:

1
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.

Follow up:

Could you do this in one pass?

Solution:

Since we’re going to do it in one pass, we need to find a way to create a gap of n+1. The reason is because, we will traverse to the end of the linked list, and we can use another pointer n+1 steps slower to find the node before nth node from end of list.

For example:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
/**
 * 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) {
        ListNode *dummy= new ListNode(0);
        dummy->next= head;
        ListNode* slow= dummy, *fast= dummy;
        
        for(int i= 0; i< n+1; i++){
            fast= fast->next;
        }
        //cout<<fast->val;
        while(fast){
            slow= slow->next;
            fast= fast->next;
        }
        //printf("\nslow:%d\n", slow->val);
        slow->next= slow->next->next;
        return dummy->next;
    }
};

time complexity: $O(n)$
space complexity: $O(1)$
reference: