本文共 840 字，大约阅读时间需要 2 分钟。
问题：
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.解答：
注意如果移除的元素是头元素的话，需要特别的处理，否则的话直接找到倒数第n+1个元素，移除它后面的元素即可。
代码：
class Solution {public: ListNode *removeNthFromEnd(ListNode *head, int n) { int len=0; ListNode *pf = head; while(pf != NULL) { ++len; pf = pf->next; } if(len == n) return head->next; pf = head; ListNode *pl = head; for(int i = 1; i <= n && pf->next != NULL; i++) pf = pf->next; while(pf->next != NULL) { pf = pf->next; pl = pl->next; } pl->next = pl->next->next; return head; }}; 转载地址：http://fktsi.baihongyu.com/