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.
/** * 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) { if(head==NULL||head->next==NULL)return NULL; ListNode *former=head; ListNode *later=head; ListNode *temp=NULL; int count=0; while(former->next) { count++; former=former->next; if(count>=n) { temp=later; later=later->next; } } if(temp==NULL)head=later->next; else temp->next=later->next; delete later; return head; } };
leetcode[19]Remove Nth Node From End of List
标签：
原文地址：http://www.cnblogs.com/Vae98Scilence/p/4283669.html
