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:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
//if(null==head) return null;
ListNode scan = head;
for(int i=0; i<n; i++) scan = scan.next;
ListNode handle = new ListNode(0);
handle.next = head;
while(scan!=null){
handle = handle.next;
scan = scan.next;
}
if(handle.next==head) return head.next;
ListNode tmp = handle.next;
if(tmp!=null) handle.next = tmp.next;
return head;
}
}