README_EN.md

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

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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 and 10^4 nodes.
• The value of each node in the linked list will be in the range [1, 10^6].
• 1 <= m,n <= 1000

Solutions

Python3

# 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 i in range(m - 1):
                if pre:
                    pre = pre.next
            if pre is None:
                return head
            cur = pre
            for i in range(n):
                if cur:
                    cur = cur.next
            pre.next = None if cur is None else cur.next
            pre = pre.next
        return head

Java

/**
 * 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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