Today I have done leetcode's December LeetCoding Challenge with cpp.

December LeetCoding Challenge 6

Description

Odd Even Linked List

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

Example 1:

Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]

Example 2:

Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

Constraints:

• The number of nodes in the linked list is in the range [0, 104].
• -106 <= Node.val <= 106

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  ListNode* oddEvenList(ListNode* head) {
    ListNode *oddHead = new ListNode();
    ListNode *evenHead = new ListNode();
    ListNode *oddCur = oddHead;
    ListNode *evenCur = evenHead;
    while(head) {
      oddCur->next = head;
      oddCur = oddCur->next;
      head = head->next;
      evenCur->next = head;
      evenCur = evenCur->next;
      if(head) head = head->next;
    }
    oddCur->next = evenHead->next;
    return oddHead->next;
  }
};

// Accepted
// 70/70 cases passed (8 ms)
// Your runtime beats 92.49 % of cpp submissions
// Your memory usage beats 29.12 % of cpp submissions (10.6 MB)