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1474. Delete N Nodes After M Nodes of a Linked List

Description

You are 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.

 

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 the 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.

 

Constraints:

  • The number of nodes in the list is in the range [1, 104].
  • 1 <= Node.val <= 106
  • 1 <= m, n <= 1000

 

Follow up: Could you solve this problem by modifying the list in-place?

Solutions

  • /**
     * 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 pre = head;
            while (pre != null) {
                for (int i = 0; i < m - 1 && pre != null; ++i) {
                    pre = pre.next;
                }
                if (pre == null) {
                    return head;
                }
                ListNode cur = pre;
                for (int i = 0; i < n && cur != null; ++i) {
                    cur = cur.next;
                }
                pre.next = cur == null ? null : cur.next;
                pre = pre.next;
            }
            return head;
        }
    }
    
  • /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode() : val(0), next(nullptr) {}
     *     ListNode(int x) : val(x), next(nullptr) {}
     *     ListNode(int x, ListNode *next) : val(x), next(next) {}
     * };
     */
    class Solution {
    public:
        ListNode* deleteNodes(ListNode* head, int m, int n) {
            auto pre = head;
            while (pre) {
                for (int i = 0; i < m - 1 && pre; ++i) {
                    pre = pre->next;
                }
                if (!pre) {
                    return head;
                }
                auto cur = pre;
                for (int i = 0; i < n && cur; ++i) {
                    cur = cur->next;
                }
                pre->next = cur ? cur->next : nullptr;
                pre = pre->next;
            }
            return head;
        }
    };
    
  • # Definition for singly-linked list.
    # class ListNode:
    #     def __init__(self, val=0, next=None):
    #         self.val = val
    #         self.next = next
    class Solution:
        def deleteNodes(self, head: ListNode, m: int, n: int) -> ListNode:
            pre = head
            while pre:
                for _ in range(m - 1):
                    if pre:
                        pre = pre.next
                if pre is None:
                    return head
                cur = pre
                for _ in range(n):
                    if cur:
                        cur = cur.next
                pre.next = None if cur is None else cur.next
                pre = pre.next
            return head
    
    
  • /**
     * Definition for singly-linked list.
     * type ListNode struct {
     *     Val int
     *     Next *ListNode
     * }
     */
    func deleteNodes(head *ListNode, m int, n int) *ListNode {
    	pre := head
    	for pre != nil {
    		for i := 0; i < m-1 && pre != nil; i++ {
    			pre = pre.Next
    		}
    		if pre == nil {
    			return head
    		}
    		cur := pre
    		for i := 0; i < n && cur != nil; i++ {
    			cur = cur.Next
    		}
    		pre.Next = nil
    		if cur != nil {
    			pre.Next = cur.Next
    		}
    		pre = pre.Next
    	}
    	return head
    }
    

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