LeetCode 19. Remove Nth Node From End of List

Description

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

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.

Follow up:

Could you do this in one pass?

Explanation

We introduce a preDelete pointer first and place it at the beginning of the linked list.

1. Move head pointer first so that the distance between preDelete and head pointer would be N nodes.
2. Move head and preDelete pointer together until head is reaching to the end.
3. Modify preDelete pointing node.next to the next node of the delete node.

Java Solution

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

Python Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:        
        slow = head
        fast = head
        
        for i in range(0, n):
            fast = fast.next
            if fast == None:
                head = head.next
                return head
        
        while fast.next != None:            
            fast = fast.next
            slow = slow.next
        
        slow.next = slow.next.next
        
        return head

• Time complexity: O(L). The algorithm makes one traversal of the list of L nodes.
• Space complexity: O(1). We only used constant extra space.