LEETCODE_328 Odd Even Linked List

最新推荐文章于 2025-12-18 17:00:31 发布

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本文介绍了一种解决LeetCode中“奇偶链表”问题的高效算法。该算法使用两个指针分别处理奇数节点和偶数节点，并最终将它们合并。此方法实现了O(1)的空间复杂度和O(nodes)的时间复杂度。

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:
Given 1->2->3->4->5->NULL,

return 1->3->5->2->4->NULL.

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这道题是leetcode的easy题目，解法也很简单，希望和大家分享一下。基本上就two pointers. 一个用于连接所有的偶数nodes，一个用于连接所有的基数nodes。最后把尾he头接上就好啦~~

class Solution {
public:
ListNode* oddEvenList(ListNode* head) {
if(head == nullptr || head->next == nullptr) // If linkedList contains less than 2 nodes, return head
return head;
ListNode* first = head;
ListNode* second = head->next;
ListNode* temp = second;
while(second && second->next){ // 每次偶数的list连接一个偶数node，基数list也连接一个基数node。
first->next = second->next;
first = first->next;
second->next = first->next;
second = second->next;
}
first->next = temp; //尾部和头部相连
return head;
}
};