将链表按照index，先odd index再odd index重排
空间复杂度O(1)
时间复杂度O(n)
主要是考虑到 odd=odd->next->next后，原始list中odd后断开，找不到odd->next,故一次两个推进，odd和even
ListNode* odd = head; ListNode* even = head->next; ListNode* evenhead = even; if(odd!=null){ while(even!=NULL&&even->next!=NULL){ odd->next = odd->next->next; even->next = even->next->next; odd = odd->next; even = even->next; } odd->next = evenhead; }
原文：https://www.cnblogs.com/lqerio/p/11701488.html