leetcode 328. Odd Even Linked List

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本文介绍了一种在不额外使用空间的情况下对单链表进行奇偶节点重组的方法。该算法确保了奇数位置节点和偶数位置节点分别连续排列，并保持原有的相对顺序。通过遍历链表并调整指针连接实现目标结构，最终达到所需的时间复杂度为O(n)。

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

/**
 * 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 || !head->next || !head->next->next)
            return head;
        
        ListNode* last = head;
        
        while (last->next)
        {
            last = last->next;
        }
        
        ListNode* p = last;
        ListNode* begin = head;
        ListNode* q;
        while (begin != last)
        {
            q = begin->next;
            if (q == last)
                last = last->next;
            begin->next = q->next;
            q->next = NULL;
            p->next = q;
            p = q;
            begin = begin->next;
        }
        return head;    
    }
};