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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.
Use two pointers, slow and fast. Fast pointer firstly move n steps. Then, when fast pointer reaches end, slow pointer reaches the node before deleted node.
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public ListNode removeNthFromEnd(ListNode head, int n) { 11 ListNode fast = head, slow = head; 12 if (head == null) 13 return head; 14 while (n > 0) { 15 fast = fast.next; 16 n--; 17 } 18 // If remove the first node 19 if (fast == null) { 20 head = head.next; 21 return head; 22 } 23 while (fast.next != null) { 24 fast = fast.next; 25 slow = slow.next; 26 } 27 slow.next = slow.next.next; 28 return head; 29 } 30 }
Remove Nth Node From End of List 解答
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4812038.html
