[NeetCode 150] Remove Node From End of Linked List

Remove Node From End of Linked List

You are given the beginning of a linked list head, and an integer n.

Remove the nth node from the end of the list and return the beginning of the list.

Example 1:

Input: head = [1,2,3,4], n = 2

Output: [1,2,4]

Example 2:

Input: head = [5], n = 1

Output: []

Example 3:

Input: head = [1,2], n = 2

Output: [2]

Constraints:

The number of nodes in the list is sz.

1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

Solution

The key is to find the target node. To do this, we can use two pointers algorithm. Let first point steps $n$ times first, and then both pointers step together until the first one meets the end.

Code

An additional head node may be helpful. Here I just specially judge the situation that the node to be removed is the head node.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        p1, p2 = head, head
        pre = ListNode(0)
        step = 0
        while p1.next:
            p1 = p1.next
            step += 1
            if step >= n:
                pre = p2
                p2 = p2.next
        print(p1.val, p2.val, pre.val)
        if p2 == head:
            return head.next
        pre.next = p2.next
        return head