https://leetcode.com/problems/odd-even-linked-list/description/ 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 ...

1  public ListNode oddEvenList(ListNode head) { 2         if (head == null || head.next == null) return head; 3         ListNode oddCurr = head; 4         ListNode evenHead = head.next ; 5         ListNode evenCurr = head.next; 6         /* 7          1->2->3->4->5->NULL 8          co ce 9             eh10          1->3     2->411               ->12          * */13         14         /*15         even is always one node ahead so:16          evenCurr != null == oddCurr.next != null17          evenCurr.next != null == oddCurr.next.next != null18 19         重要！when evenCurr is at 4, evenCurr.next = 5 != null20         */21         while (evenCurr != null && evenCurr.next != null){22             oddCurr.next = oddCurr.next.next ;23             oddCurr = oddCurr.next;24             //4->5.next=null 4->null25             evenCurr.next = evenCurr.next.next ;26             evenCurr = evenCurr.next;27         }28         //奇数结尾指向偶数的头29         oddCurr.next = evenHead ;30         return head;31     }

千万注意，不要遍历两遍 什么 while(oddcurr)  while(evencurr)  因为underline 的链表是一个，所以遍历第一遍之后就已经把 next 的关系搞混了， 后面那个遍历就不准确了。