Question
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.
Solution -- One pass
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.
/**
* 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) {
ListNode fast = head, slow = head;
if (head == null)
return head;
while (n > 0) {
fast = fast.next;
n--;
}
// If remove the first node
if (fast == null) {
head = head.next;
return head;
}
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return head;
}
}