328. Odd Even Linked List
- Hardness: \(\color{orange}\textsf{Medium}\)
- Ralated Topics:
Linked List
一、題目
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, 10^4]. -10^6 <= Node.val <= 10^6
二、分析
- 分別將奇偶位置的鏈表分開相接之後再將其頭尾串連。
- 注意因為我們要拿奇數的尾接偶數的頭，故要注意 iterate 的終止條件是
!even || !even->next
三、解題
1. Linked List
- Time complexity: \(O(n)\)
- Space complexity: \(O(1)\)
ListNode* oddEvenList(ListNode* head) {
if (!head) return NULL;
ListNode* head2 = head->next;
ListNode* odd = head;
ListNode* even = head2;
while (even && even->next) {
odd->next = odd->next->next;
even->next = even->next->next;
odd = odd->next;
even = even->next;
}
odd->next = head2;
return head;
}