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 …

原題鏈接：https://leetcode.com/problems/odd-even-linked-list/

思路
奇節點的下一個節點恰好就是偶節點的下一個節點（1的下一個節點應該是2的下一個節點，也就是3）;
而偶節點的下一個節點也恰好是奇節點是下一個節點
所以只需要odd even兩個循環往下遍歷就可以了，最後把奇鏈表的尾部指向偶鏈表的頭。

代碼（C++）

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