[LeetCode] 19. Remove Nth Node From End of List

Posted 2020-08-12

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.

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* removeNthFromEnd(ListNode* head, int n) {
12         ListNode *dummy = new ListNode(0);
13         dummy->next = head;
14         ListNode *ret = NULL;
15         ListNode *prev = dummy;
16         
17         ListNode *p1 = head;
18         ListNode *p2 = head;
19         int counter = 0;
20         
21         while (p2){
22             p2 = p2->next;
23             counter++;
24             if (counter == n){
25                 break;
26             }
27         }
28         
29         if (counter < n){
30             delete dummy;
31             return head;
32         }
33         
34         while(p2){
35             p1 = p1->next;
36             p2 = p2->next;
37             prev = prev->next;
38         }
39         
40         prev->next = p1->next;
41         
42         ret = dummy->next;
43         delete dummy;
44         return ret;
45     }
46 };