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1474. Delete N Nodes After M Nodes of a Linked List πŸ”’

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

Thinking

Repeatedly keep \(m\) nodes and drop \(n\). Advance \(\textit{pre}\) over the kept segment, skip \(n\) nodes with \(\textit{cur}\), then relink \(\textit{pre.next}\). Stop early if the list runs out.

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 \(O(1)\).

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# 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

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/**
 * 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;
    }
}

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/**
 * 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;
    }
};

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/**
 * 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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/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function deleteNodes(head: ListNode | null, m: number, n: number): ListNode | null {
    let pre = head;
    while (pre) {
        for (let i = 0; i < m - 1 && pre; ++i) {
            pre = pre.next;
        }
        if (!pre) {
            break;
        }
        let cur = pre;
        for (let i = 0; i < n && cur; ++i) {
            cur = cur.next;
        }
        pre.next = cur?.next || null;
        pre = pre.next;
    }
    return head;
}

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