Leetcode - 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.

实现的代码关键是从算法的角度来讨论文件的解决思路。

下面代码在实际应用中，会导致删除节点内存没有释放，从而导致内存泄露。

解决方法，只要在修改 *pNode 之前，备份这个值， 在赋值之后，delete 备份的指正既可。

       struct ListNode {           int val;           ListNode *next;           ListNode(int x) : val(x), next(NULL) {}      };

ListNode* removeNthFromEnd(ListNode* head, int n) {       ListNode **pNode = (ListNode**) &head;       ListNode *first = head;       for(int i = 0 ; i < n  ; i ++ ) { first = first->next ; }              while(first != NULL) {           first = first->next;           pNode = (ListNode**) &(*pNode)->next;       }       *pNode = (*pNode)->next;              return head;    }