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1 题目
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.
2 思路
题目要求遍历一遍搞定，那么，就可以用两个指针，先第一个先往前走N个，然后再一起走，最后，移除后一个元素。自己做出来的。。
3 代码
public ListNode removeNthFromEnd(ListNode head, int n) { ListNode node = head; while(n>0 && node!=null){ node = node.next; n--; } ListNode start = head; if(node != null){ while(node.next != null){ node = node.next; start = start.next; } }else{ /* will remove the first object */ start = start.next; return start; } start.next = start.next.next; return head; }
[leetcode 19] Remove Nth Node From End of List
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原文地址：http://www.cnblogs.com/lingtingvfengsheng/p/4779323.html
