LeetCode之Middle of the Linked List（Kotlin）

问题： Given a non-empty, singly linked list with head node head, return a middle node of linked list. If there are two middle nodes, return the second middle node.

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3.  (The judge's serialization of this node is [3,4,5]). Note that we returned a ListNode object ans, such that: ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL. Example 2: Input: [1,2,3,4,5,6] Output: Node 4 from this list (Serialization: [4,5,6]) Since the list has two middle nodes with values 3 and 4, we return the second one. Note: The number of nodes in the given list will be between 1 and 100. 复制代码

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方法： 经过递归调用保证不丢失前向node，计算当前node的前向index和后向index，当前向index等于后向index或后向index加1时当前node即为中间node。

具体实现：

class MiddleOfTheLinkedList {
    // Definition for singly-linked list.
    class ListNode(var `val`: Int = 0) {
        var next: ListNode? = null
    }

    fun middleNode(head: ListNode?): ListNode? {
        return middleNode(head, 1)
    }

    fun getBackIndex(head: ListNode?): Int {
        if (head == null) {
            return 0
        }
        return getBackIndex(head.next) + 1
    }

    fun middleNode(head: ListNode?, frontIndex : Int): ListNode? {
        if (head == null) {
            return null
        }
        if (frontIndex == getBackIndex(head)
                || frontIndex - 1 == getBackIndex(head)) {
            return head
        } else{
            return middleNode(head.next, frontIndex + 1)
        }
    }
}

fun main(args: Array<String>) {

}
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有问题随时沟通

具体代码实现能够参考Github