328. Odd Even Linked List

1.Question

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.

PS:先把单节点连接在一起，再把双节点连接在一起，最后把双节点拼接到但节点的后面。

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 ...

2.Code

/**
 * 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 ||head->next == NULL) return head;
            ListNode *odd_pre = head, *even_pre = head->next, *even_str = head->next;
	    while(even_pre->next != NULL && even_pre->next->next != NULL)
        {
		odd_pre->next = even_pre->next;
		odd_pre = odd_pre->next;
		even_pre->next = odd_pre->next;
		even_pre = even_pre->next;
        }
	    if(even_pre->next == NULL) odd_pre->next = even_str;
	    else
	    {
		odd_pre->next = even_pre->next;
		odd_pre->next->next = even_str;
		even_pre->next = NULL;
	    }
        return head;
    }
};

3.Note

a. 在判断 if(head != NULL) 时，可以直接写成if(head) ;

b.在定义两个结构体指针时，要写清楚 ListNode *odd_pre = head, *even_pre = head->next; 不要写成ListNode *odd_pre = head, even_pre = head->next;  造成这种错误。