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.
很典型的2 pointer题目。先前进n, 然后当快的pointer到tail时候，慢的正好到要删除的前一个。
public ListNode RemoveNthFromEnd(ListNode head, int n) { ListNode slow = head; ListNode fast = head; ListNode sentinel = new ListNode(-1); sentinel.next = head; //first get the n distance between slow and fast, as it always valid, no need for valid check for(int i =0;i<n;i++) { fast = fast.next; } if(fast == null)//the first nood { sentinel.next = head.next; return sentinel.next; } while(fast.next != null) { slow = slow.next; fast = fast.next; } slow.next = slow.next.next; return head; }
19. Remove Nth Node From End of List
原文：http://www.cnblogs.com/renyualbert/p/5867552.html