LeetCode //C - 328. Odd Even Linked List

328. 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:

Example 2:

Constraints:

• The number of nodes in the linked list is in the range [ 0 , 1 0 4 ] [0, 10^4] [0,104].
• − 1 0 6 < = N o d e . v a l < = 1 0 6 -10^6 <= Node.val <= 10^6 −106<=Node.val<=106

From: LeetCode
Link: 328. Odd Even Linked List

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Solution:

Ideas：

This function works by first checking if the head is NULL. If it’s not, it creates two pointers, odd and even, which point to the first and second nodes of the list, respectively. evenHead stores the head of the even list.

The loop continues until there are no more even nodes or even nodes with a next node. Inside the loop, the odd nodes are connected to the next odd node, and similarly, the even nodes are connected to the next even node. After the end of the loop, the odd list and the even list are connected.

The final list starts with the odd nodes followed by the even nodes, as shown in your images.

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* oddEvenList(struct ListNode* head) {
    if (head == NULL) return NULL;
    
    struct ListNode *odd = head;
    struct ListNode *even = head->next;
    struct ListNode *evenHead = even;

    while (even != NULL && even->next != NULL) {
        odd->next = odd->next->next;
        even->next = even->next->next;
        odd = odd->next;
        even = even->next;
    }
    
    odd->next = evenHead;
    return head;
}