#Leetcode# 328. Odd Even Linked List

最新推荐文章于 2025-09-11 18:44:42 发布

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最新推荐文章于 2025-09-11 18:44:42 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zlrrrr/p/10293517.html

本文介绍了一种在链表中将所有奇数节点按顺序排列在所有偶数节点之前的算法实现。该算法遵循节点编号而非节点值进行排序，且在输入中保持了偶数和奇数组内部的相对顺序不变。文章提供了C++代码示例，展示了如何通过在O(1)空间复杂度和O(nodes)时间复杂度下操作，实现这一链表重构。

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 ...

代码：

/**
 * 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) return NULL;
        ListNode *odd = head;
        ListNode *even = head -> next;
        ListNode *mid = even;
        
        while(odd && even && odd -> next && even -> next) {
            odd -> next = odd ->next -> next;
            even -> next = even -> next -> next;
            odd = odd -> next;
            even = even -> next;
        }
        odd -> next = mid;
        return head;
    }
};

转载于:https://www.cnblogs.com/zlrrrr/p/10293517.html