【Leetcode】【Easy】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.

普通青年解法：

设置三个指针A\B\C，指针A和B间隔n-1个结点，B在前A在后，用指针A去遍历链表直到指向最后一个元素，则此时指针B指向的结点为要删除的结点，指针C指向指针B的pre，方便删除的操作。

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 *target = head;
         ListNode *target_pre = NULL;
         ListNode *findthelast = head;
 
         if (head == NULL)
             return head;
 
         while(--n>)
             findthelast = findthelast->next;
 
         while (findthelast->next) {
             target_pre = target;
             target = target->next;
             findthelast = findthelast->next;
         }
 
         if (target_pre == NULL) {
             head = target->next;
             return head;
         }
 
         target_pre->next = target->next;
 
         return head;
     }
 };

文艺智慧青年解法：

（需要更多理解）

 class Solution
 {
 public:
     ListNode* removeNthFromEnd(ListNode* head, int n)
     {
         ListNode** t1 = &head, *t2 = head;
         for(int i = ; i < n; ++i)
         {
             t2 = t2->next;
         }
         while(t2->next != NULL)
         {
             t1 = &((*t1)->next);
             t2 = t2->next;
         }
         *t1 = (*t1)->next;
         return head;
     }
 };

附录：

C++二重指针深入