[lintcode easy]Remove Nth Node From End of List

Remove Nth Node From End of List

 

Given a linked list, remove the nth node from the end of list and return its head.

 

 

Example

Given linked list: 1->2->3->4->5->null, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5->null.

Note

The minimum number of nodes in list is n.

Challenge

O(n) time

 

////same as return Nth Node to End

////Using two pointer to track node, keep their distance as N step.

///If n is the the first node, return head.next;  else ,return head;

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param n: An integer.
     * @return: The head of linked list.
     */
    ListNode removeNthFromEnd(ListNode head, int n) {
        // write your code here
        if(head==null || n<0) return head;
        ListNode fast=head;
        ListNode slow=new ListNode(0);
        slow.next=head;
        
        while(n--!=0)
        {
            fast=fast.next;
        }
        if(fast==null)
        {
            return head.next;
        }
        while(fast!=null)
        {
            fast=fast.next;
            slow=slow.next;
            
        }
        
        ListNode next=slow.next;
        slow.next=next.next;
        
        return head;
    }
}