题目来源:328. Odd Even Linked List
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:
Given1->2->3->4->5->NULL
,
return1->3->5->2->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 …Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.Subscribe to see which companies asked this question
方案1:单一链表内指针操作,实现奇偶元素位置调整
1 | struct ListNode* oddEvenList(struct ListNode* head) { |
这个思路挺有意思,画图说明下:
方案2:两个链表头分别指向基数链表和偶数链表,最终连接奇偶链表
1 | ListNode* oddEvenList(ListNode* head) { |