【LeetCode】19. Remove Nth Node From End of List

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.

两个指针相隔n-1，当一个指针走到末尾了，后一个指针就是要删除的节点。

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