328. Odd Even Linked List

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本文介绍如何在不使用额外空间的情况下，将链表中奇数节点与偶数节点进行分组，保持相对顺序不变。

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【题目】

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 ...

【思路】

其实是道很简单的链表题，不过前面两个月一直没做题，明显手生了，一开始竟然思路卡住了=。=

主要思路是设置两个指针，一个指针指向odd结点，一个指向even结点，group之后odd结点的next必然是之前even结点的next，而even结点的next必然是本来是even结点的next（即odd结点）的next。

【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==NULL) return NULL;
        if(head->next==NULL) return head;
        ListNode* oddH=head;
        ListNode* evenH=head->next;
        ListNode* odd=oddH;
        ListNode* even=evenH;
        while(odd&&even&&even->next){
            odd->next=even->next;
            odd=odd->next;
            even->next=odd->next;
            even=even->next;
        }
        odd->next=evenH;
        return oddH;
    }
};