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.
方法： 通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>) {
}
有问题随时沟通
