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

Solution 1: Simulation

We can simulate the entire deletion process. First, use a pointer $\textit{pre}$ to point to the head of the linked list, then traverse the linked list, moving $m - 1$ steps. If $\textit{pre}$ is null, it means the number of nodes from the current node is less than $m$, so we directly return the head. Otherwise, use a pointer $\textit{cur}$ to point to $\textit{pre}$, then move $n$ steps. If $\textit{cur}$ is null, it means the number of nodes from $\textit{pre}$ is less than $m + n$, so we directly set the $\textit{next}$ of $\textit{pre}$ to null. Otherwise, set the $\textit{next}$ of $\textit{pre}$ to the $\textit{next}$ of $\textit{cur}$, then move $\textit{pre}$ to its $\textit{next}$. Continue traversing the linked list until $\textit{pre}$ is null, then return the head.

The time complexity is $O(n)$, where $n$ is the number of nodes in the linked list. The space complexity is $O(1)$.

Python3JavaC++GoTypeScript

# 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.
 * 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.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }