Problem Statement
Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
Example 1:
Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
Example 2:
Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL
Note:
- The relative order inside both the even and odd groups should remain as it was in the input.
- The first node is considered odd, the second node even and so on ...
Solution Thought Process
For this problem, we have to create two linked list:
-
OddLinkedList- for containing odd index elements -
EvenLinkedList- for containing even index elements
We iterate over the linked list, we initialize a counter to 1. Whenever -
- The counter is divisible by 2, we add the node to the
OddLinkedList - The counter is divisible by 1, we add the node to the
EvenLinkedList
After that, we merge the tail of OddLinkedList to the EvenLinkedList head. And in the end, we make the tail of EvenLinkedList point to NULL to make sure that the linked list has ended in NULL.
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* oddEvenList(ListNode* head) {
ListNode* oddLinkedList = new ListNode();
ListNode* evenLinkedList = new ListNode();
ListNode* oddIterator = oddLinkedList;
ListNode* evenIterator = evenLinkedList;
ListNode* iterator = head;
int counter = 1;
while(iterator)
{
if(counter % 2)
{
oddIterator->next = iterator;
iterator = iterator->next;
oddIterator = oddIterator->next;
oddIterator->next = NULL;
}
else {
evenIterator->next = iterator;
iterator = iterator->next;
evenIterator = evenIterator->next;
evenIterator->next = NULL;
}
counter++;
}
oddIterator->next = evenLinkedList->next;
evenIterator->next = NULL;
return oddLinkedList->next;
}
};
Complexity:
Time Complexity: O(n)
Space Complexity: O(1)
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