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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; } }
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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. # 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. * 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; }