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.
题解：关键点在于把倒数第n位移除变成正数的来看，第一次遍历记录链表长度，第二次遍历到要删除的结点位置。
/** * 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(n==0) { return head; } int cnt=0; ListNode* p=head; ListNode* ret=head; while(p!=NULL) { cnt++; p=p->next; } int pos=cnt-n-1; if(pos<0) { return head->next; } while(pos>0) { head=head->next; pos--; } head->next=head->next->next; return ret; } };
Leetcode-19 Remove Nth Node From End of List
原文：http://www.cnblogs.com/fengxw/p/6082835.html