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.

Two passes solution, O(n) time, O(1) space

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/**
 * 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) {
        int length = 0;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode first = head;
        while (first != null){
            length++;
            first = first.next;
        }
        length -= n;
        first = dummy;
        while (length > 0){
            first = first.next;
            length--;
        }
        first.next = first.next.next;
        return dummy.next;
    }
}

One pass solution, O(n) time, O(1) space

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/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        for (int i = 0; i <= n; i++){
            fast = fast.next;
        }
        // if the two pointers are n nodes away, excluding themselves, then the while condition is "fast != null"
        // if the two pointers are n - 1 nodes away, the condition should be "first.next != null"
        while (fast != null){
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }
}