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?
# 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
/**
* 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;
}
}