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.
Solution:
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode removeNthFromEnd(ListNode head, int n) { 14 if (head==null) return head; 15 16 ListNode first = head; 17 ListNode second = head; 18 for (int i=0;i<n;i++) 19 first = first.next; 20 21 if (first==null) return second.next; 22 23 while (first.next!=null){ 24 second = second.next; 25 first = first.next; 26 } 27 28 ListNode temp = second.next.next; 29 second.next = temp; 30 return head; 31 } 32 }
Leetcode-Remove Nth Node From End of List
原文：http://www.cnblogs.com/lishiblog/p/4130201.html