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.
思路：先从头遍历等到长度length，在从头遍历，根据当前结点次序与长度length和n的关系确定要删除的结点。
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 *p=head,*q; 13 int length=0; 14 while(p!=NULL) 15 { 16 length++; 17 p=p->next; 18 } 19 p=head; 20 head=q; 21 q->next=p; 22 while(p!=NULL) 23 { 24 if(length==n) 25 { 26 q->next=p->next; 27 break; 28 } 29 else 30 { 31 q=p; 32 p=p->next; 33 length--; 34 } 35 } 36 return head->next; 37 } 38 };
LeetCode:Remove Nth Node From End of List
原文：http://www.cnblogs.com/levicode/p/3974740.html