LeetCode之Odd Even Linked List（Kotlin）

問題： 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 ...
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方法： 雙指針法，一個用來指向奇數鏈，一個用來指向偶數鏈，若是奇數鏈指針和偶數鏈指針有下一個的話就執行循環，奇數鏈指向偶數鏈的下一個，而後奇數鏈向下滑一個位置，同時偶數鏈指向奇數鏈的下一個，而後偶數鏈向下滑一個，最後奇數鏈的最後一個指向偶數鏈的根節點，最後返回奇數鏈的根節點。

具體實現：

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>) {
}
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有問題隨時溝通

具體代碼實現能夠參考Github