1 Description #
source: https://leetcode.com/problems/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, and 5, the middle nodes are 0, 1, 1, 2, and 2, 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
2 Solution #
/**
* 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* deleteMiddle(ListNode* head) {
// Time complexity: O(1.5N) = O(N). N is the size of linked list
// Space complexity: O(1)
if(!head){
return head;
}
int size = 0;
ListNode* first = head;
while(first){
size++;
first = first->next;
}
if(1 == size){
return nullptr;
}
int middle = size / 2;
int index = 0;
first = head;
while(first){
if(index == middle - 1){
// the prev node of middle node
first->next = first->next->next;
break;
}
index ++;
first = first->next;
}
return head;
}
};