LeetCode之Odd Even Linked List（Ko

问题：
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 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->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 ...

方法：
双指针法，一个用来指向奇数链，一个用来指向偶数链，如果奇数链指针和偶数链指针有下一个的话就执行循环，奇数链指向偶数链的下一个，然后奇数链向下滑一个位置，同时偶数链指向奇数链的下一个，然后偶数链向下滑一个，最后奇数链的最后一个指向偶数链的根节点，最后返回奇数链的根节点。

具体实现：

class OddEvenLinkedList {
    class ListNode(var `val`: Int) {
        var next: ListNode? = null
    }

    fun oddEvenList(head: ListNode?): ListNode? {
        val oddRoot = head
        val evenRoot = head?.next
        var oddCur = oddRoot
        var evenCur = evenRoot
        while (oddCur?.next != null || evenCur?.next != null) {
            oddCur?.next = evenCur?.next
            if (oddCur?.next != null) {
                oddCur =  oddCur.next
            }
            evenCur?.next = oddCur?.next
            evenCur = evenCur?.next
        }
        oddCur?.next = evenRoot
        return oddRoot
    }
}

fun main(args: Array<String>) {
}

有问题随时沟通

具体代码实现可以参考Github