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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.
Linked List Two Pointers
/** * 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) { ListNode *temp = head; int len=0,m; while(temp){ ++len; temp=temp->next; } if(n==len) return head->next; m=len-n; ListNode *temp1=head; while(m>1){ temp1=temp1->next; --m; } temp1->next=temp1->next->next; return head; } };
删除指定节点Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/li303491/p/4244796.html
