Leetcode刷题总结: 328. Odd Even Linked List

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 ...

题解：

简单思路，对链表计数，然后计数为奇数就添加到奇数nodelist后面，否则添加到偶数list后面；

需要注意的是，对于偶数list的最后一个元素，需要将其next置为NULL;

/**
 * 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) {
        ListNode *podd = new ListNode(-1);
        ListNode *peven = new ListNode(-1);
        ListNode *pevenHead = peven;
        ListNode *poddHead = podd;
        ListNode *pCur = head;
        int cnt = 1;
        while (pCur!=NULL) {
            if (cnt & 0x1) {
                podd->next = pCur;
                podd = pCur;
            }else {
                peven->next = pCur;
                peven = pCur;
            }
            pCur = pCur->next;
            ++cnt;
        }
        peven->next = NULL;
        podd->next = pevenHead->next;
        delete pevenHead;
        delete poddHead;
        return head;

    }
};

　思路二：

上面这个思路的效率不行，看了一下Solution里面的最快的答案，是直接循环处理了奇数偶数的next赋值，确实更好；