2022-10-14 Daily-Challenge
Today I have done leetcode's October LeetCoding Challenge with cpp
.
October LeetCoding Challenge 14
Description
Delete the Middle Node of a Linked List
You are given the head
of a linked list. Delete the middle node, and return the head
of the modified linked list.
The middle node of a linked list of size n
is the ⌊n / 2⌋th
node from the start using 0-based indexing, where ⌊x⌋
denotes the largest integer less than or equal to x
.
- For
n
=1
,2
,3
,4
, and5
, the middle nodes are0
,1
,1
,2
, and2
, respectively.
Example 1:
Input: head = [1,3,4,7,1,2,6]
Output: [1,3,4,1,2,6]
Explanation:
The above figure represents the given linked list. The indices of the nodes are written below.
Since n = 7, node 3 with value 7 is the middle node, which is marked in red.
We return the new list after removing this node.
Example 2:
Input: head = [1,2,3,4]
Output: [1,2,4]
Explanation:
The above figure represents the given linked list.
For n = 4, node 2 with value 3 is the middle node, which is marked in red.
Example 3:
Input: head = [2,1]
Output: [2]
Explanation:
The above figure represents the given linked list.
For n = 2, node 1 with value 1 is the middle node, which is marked in red.
Node 0 with value 2 is the only node remaining after removing node 1.
Constraints:
- The number of nodes in the list is in the range
[1, 10^5]
. 1 <= Node.val <= 10^5
Solution
auto speedup = [](){
cin.tie(nullptr);
cout.tie(nullptr);
ios::sync_with_stdio(false);
return 0;
}();
class Solution {
public:
ListNode* deleteMiddle(ListNode* head) {
ListNode *newHead = new ListNode(0, head);
ListNode *fast = newHead;
ListNode *slow = newHead;
ListNode *cur = newHead;
while(fast->next && fast->next->next) {
fast = fast->next;
if(fast->next) {
fast = fast->next;
}
slow = slow->next;
}
slow->next = slow->next->next;
return newHead->next;
}
};
// Accepted
// 70 / 70 cases passed (1043 ms)
// Your runtime beats 97.62 % of cpp submissions
// Your memory usage beats 20.46 % of cpp submissions (294.9 MB)