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.

Remove Nth Node From End of List.cpp

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(head == NULL) {
            return head;
        }
        
        ListNode tmpHead(-1);
        tmpHead.next = head;
        ListNode* cur = &tmpHead;
        ListNode* forward = cur;
        while(n--) {
            forward = forward->next;
        }
        
        while(forward->next != NULL) {
            forward = forward->next;
            cur = cur->next;
        }
        
        ListNode* deleted = cur->next;
        cur->next = deleted->next;
        return tmpHead.next;
        
    }
};