Leetcode 19. Remove Nth Node From End of List (Easy) (cpp)

Leetcode 19. Remove Nth Node From End of List (Easy) (cpp)

Tag: Linked List, Two Pointers

Difficulty: Easy

/*

19. Remove Nth Node From End of List (Easy)

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.

*/
/**
* 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 (n < 1 || head == NULL) return head;
ListNode *p = head, *q = head, *q_prev = NULL;
for (int i = 0; i < n - 1; i++)
p = p -> next;
while (p -> next) {
p = p -> next;
q_prev = q;
q = q -> next;
}
if (q_prev == NULL) return head -> next;
else {
q_prev -> next = q -> next;
return head;
}
}
};