LeetCode – Remove Nth Node From End of List (Java)

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, the result is 1->2->3->5.

Java Solution 1 - Naive Two Passes

Calculate the length first, and then remove the nth from the beginning.

public ListNode removeNthFromEnd(ListNode head, int n) {
    if(head == null)
        return null;
 
    //get length of list
    ListNode p = head;
    int len = 0;
    while(p != null){
        len++;
        p = p.next;
    }
 
    //if remove first node
    int fromStart = len-n+1;
    if(fromStart==1)
        return head.next;
 
    //remove non-first node    
    p = head;
    int i=0;
    while(p!=null){
        i++;
        if(i==fromStart-1){
            p.next = p.next.next;
        }
        p=p.next;
    }
 
    return head;
}

Java Solution 2 - One Pass

Use fast and slow pointers. The fast pointer is n steps ahead of the slow pointer. When the fast reaches the end, the slow pointer points at the previous element of the target element.

public ListNode removeNthFromEnd(ListNode head, int n) {
    if(head == null)
        return null;
 
    ListNode fast = head;
    ListNode slow = head;
 
    for(int i=0; i<n; i++){
        fast = fast.next;
    }
 
    //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;
}
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  • gao can

    Treat it like a circle, but I think two pointers is better.
    /**
    * 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 (head == null) return head;
    ListNode p = head;
    int len = 0;
    while (p != null) {
    len++;
    if (p.next != null) {
    p = p.next;
    } else {
    break;
    }
    }
    // delete head case
    if (n == len) return head.next;
    ListNode end = p;
    p.next = head;
    int k = len – n;
    while (k > 0) {
    k–;
    p = p.next;
    }
    end.next = null;
    // delete tail case
    if (n == 1) {
    p.next = null;
    } else {
    p.next = p.next.next;
    }
    return head;
    }
    }

  • Aswin Gokul

    we have to check for null in the for loop. Otherwise it’s going to give us NullPointerException

  • jason zhang

    Both solutions have the same number of walk. There are not much difference