标签：
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.
Analyse: 如果链表为空，返回NULL；如果n等于链表的长度，即删除head；否则，先找到该删除位置的前一位置，将该位置的next指向被删除位置的next，并释放被删除指针的空间。第一次没有WA或TE就AC了，mark一下~~ :)
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *removeNthFromEnd(ListNode *head, int n) { 12 ListNode *current = head; 13 int length = 0; 14 while(current){ 15 length++; 16 current = current->next; 17 } 18 19 if(length == 1) return NULL; 20 if(length == n){ 21 ListNode *temp = head; 22 head = head->next; 23 delete temp; 24 return head; 25 } 26 27 current = head; 28 for(int i = 1; i < length - n; i++) current = current->next; 29 30 ListNode *temp = current->next; 31 current->next = current->next->next; 32 delete temp; 33 34 return head; 35 } 36 };
Remove Nth Node From End of List
标签：
原文地址：http://www.cnblogs.com/amazingzoe/p/4419854.html
