Formatted question description: https://leetcode.ca/all/1474.html
1474. Delete N Nodes After M Nodes of a Linked List
Level
Easy
Description
Given the head
of a linked list and two integers m
and n
. Traverse the linked list and remove some nodes in the following way:
- Start with the head as the current node.
- Keep the first
m
nodes starting with the current node. - Remove the next
n
nodes - Keep repeating steps 2 and 3 until you reach the end of the list.
Return the head of the modified list after removing the mentioned nodes.
Follow up question: How can you solve this problem by modifying the list in-place?
Example 1:
Input: head = [1,2,3,4,5,6,7,8,9,10,11,12,13], m = 2, n = 3
Output: [1,2,6,7,11,12]
Explanation: Keep the first (m = 2) nodes starting from the head of the linked List (1 ->2) show in black nodes.
Delete the next (n = 3) nodes (3 -> 4 -> 5) show in read nodes.
Continue with the same procedure until reaching the tail of the Linked List. Head of linked list after removing nodes is returned.
Example 2:
Input: head = [1,2,3,4,5,6,7,8,9,10,11], m = 1, n = 3
Output: [1,5,9]
Explanation: Head of linked list after removing nodes is returned.
Example 3:
Input: head = [1,2,3,4,5,6,7,8,9,10,11], m = 3, n = 1
Output: [1,2,3,5,6,7,9,10,11]
Example 4:
Input: head = [9,3,7,7,9,10,8,2], m = 1, n = 2
Output: [9,7,8]
Constraints:
- The given linked list will contain between
1
and10^4
nodes. - The value of each node in the linked list will be in the range
[1, 10^6]
. 1 <= m,n <= 1000
Solution
Create a dummy node that is before the head. Starting from the dummy node, each time move forward m
steps, and then set the current node’s next node n
steps ahead. Repeat the process until the end of the list is reached. Finally, return the head of the list.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode deleteNodes(ListNode head, int m, int n) {
ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode temp = dummyHead;
while (temp != null && temp.next != null) {
for (int i = 1; i <= m && temp != null && temp.next != null; i++)
temp = temp.next;
for (int i = 1; i <= n && temp != null && temp.next != null; i++)
temp.next = temp.next.next;
}
return dummyHead.next;
}
}