leetcode 328. Odd Even Linked List

题目要求

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 ...

将一个链表中的节点按照奇数位上的节点在前，偶数位上的节点在后重新排序。这里需要注意的是节点之间的相对顺序不可以改变。即1->2->3->4不可以变为1->3->4->2，只能是1->3->2->4。

思路和代码

首先想到的就是直接新建两个链表头分别用来从原始的链表中提取奇数位上的节点和偶数位上的节点。最后将奇数链表尾连到偶数链表头即可。

    public ListNode oddEvenList(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode dummyOdd = new ListNode(0);
        ListNode dummyEven = new ListNode(0);
        ListNode cur = head,
                odd = dummyOdd,
                even = dummyEven;
        int count = 0;
        while(cur!=null){
            if(count%2==0){
                odd.next = cur;
                odd = odd.next;
                cur = cur.next;
            }else{
                even.next = cur;
                even = even.next;
                cur = cur.next;
            }
            count++;
        }
        odd.next = dummyEven.next;
        even.next = null;
        return dummyOdd.next;
    }

改进的思路在于减少额外的变量创建。这里我们其实可以借鉴双指针的思路。偶数指针一次前进两位，奇数指针一次前进两位。奇数指针odd的初始值为head，而偶数指针even的初始值为head.next。则下一个奇数值位于even.next上，此时将该奇数指针移动到even.next上之后，偶数指针的值则为odd.next。

    public ListNode oddEvenList2(ListNode head) {
        if(head==null) return head;
        ListNode odd,even,evenStart;
        odd=head;
        even=head.next;
        evenStart=even;
        
        while(even!=null && even.next!=null){
            odd.next=even.next;
            odd=odd.next;
            even.next=odd.next;
            even=even.next; 
        }
        odd.next=evenStart;
        return head;
    }

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