[LeetCode] (medium) 328. Odd Even Linked List-优快云博客

本文链接：https://blog.youkuaiyun.com/Frontier_Setter/article/details/87628688

本文介绍了一种在链表中按奇偶节点重新排列的方法，确保所有奇数节点位于偶数节点之前，同时保持各自组内原始顺序。该算法在O(1)空间复杂度和O(nodes)时间复杂度下实现，适用于LeetCode上的Odd Even Linked List问题。通过维护活动节点，即两段链表的末尾和未定位的头两个节点，实现了高效且简洁的解决方案。

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https://leetcode.com/problems/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 1:
Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
Example 2:
Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->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 ...

只要确定好活跃节点（两段的末尾和未就位的头两个节点）就很好做了。注意循环跳出的条件和一开始的特殊情况判断（如果人工加入头结点可以简化一开始的特殊情况，让代码更优美）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(head == NULL) return head;
        ListNode *first = head, *second = head->next;
        if(second == NULL) return head;
        ListNode *tem;
        while(second != NULL && second->next != NULL){
            tem = second->next;
            second->next = second->next->next;
            tem->next = first->next;
            first->next = tem;
            first = first->next;
            second = second->next;
        }
        return head;
    }
};