LeetCode题解(19)--Remove Nth Node From End of List

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

原题：

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.

思路：

考察指针的灵活使用。使用两个指针a和b就够了，中间间隔n个，一起向前移动，直到最后一个节点，然后删掉a->next。这样的算法需要注意删掉的是第一个节点的情况（也即不能是a->next了，直接head=head->next）。

我的AC代码：

 /**
  * 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 * to_remove, * p;
         p=head;
         for(int i=;i<n;i++){
             p=p->next;
         }
         if(p==NULL){
             head=head->next;
             return head;
         }
         to_remove=head;
         while(p->next!=NULL){
             p=p->next;
             to_remove=to_remove->next;
         }
         to_remove->next=to_remove->next->next;
         return head;
     }
 };