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 <= 1061 <= m, n <= 1000
Follow up: Could you solve this problem by modifying the list in-place?
Solutions#
Solution 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
}
|
