leetcode328

1、 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 …
思路大概是～把奇数列和偶数列分别拎出来，然后在连在一起，主要while判断，无论是奇数个还是偶数个节点，最终偶数指针会在奇数指针后面，即只要even和even->next都不为空就还没有结束，最终若是结束了，奇数指针连上最初的偶数指针就好～

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