[Swift]LeetCode876. 链表的中间结点 | Middle of the Linked List

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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.

---

给定一个带有头结点 head 的非空单链表，返回链表的中间结点。

如果有两个中间结点，则返回第二个中间结点。 

示例 1：

输入：[1,2,3,4,5]
输出：此列表中的结点 3 (序列化形式：[3,4,5])
返回的结点值为 3 。 (测评系统对该结点序列化表述是 [3,4,5])。
注意，我们返回了一个 ListNode 类型的对象 ans，这样：
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, 以及 ans.next.next.next = NULL.

示例 2：

输入：[1,2,3,4,5,6]
输出：此列表中的结点 4 (序列化形式：[4,5,6])
由于该列表有两个中间结点，值分别为 3 和 4，我们返回第二个结点。 

提示：

• 给定链表的结点数介于 1 和 100 之间。

---

Runtime: 4 ms

Memory Usage: 19.1 MB

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func middleNode(_ head: ListNode?) -> ListNode? {
        guard let _ = head else {
            return nil
        }
        var fastNode = head
        var slowNode = head
        while let curNode = fastNode, let tmpNode = curNode.next {
            slowNode = slowNode?.next
            fastNode = tmpNode.next
        }
        return slowNode
    }
}

---

4ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func middleNode(_ head: ListNode?) -> ListNode? {
        var list = [ListNode]()
        
        var currentNode: ListNode? = head
        while let node = currentNode {
            list.append(node)
            currentNode = node.next
        }
        
        return list[list.count / 2]
    }
}

---

4ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func middleNode(_ head: ListNode?) -> ListNode? {
        var slow:ListNode? = head
        var fast:ListNode? = head
        while (fast != nil && fast?.next != nil)
        {
            slow = slow?.next
            fast = fast?.next?.next
        }
        return slow
    }
}

---

8ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func middleNode(_ head: ListNode?) -> ListNode? {
      
    var i = head!
    var j = head!
        while (j != nil && j.next != nil){
             i = i.next!
             if j.next != nil || j.next!.next != nil{
             j = j.next!.next ?? ListNode(0)
            }
        }
        return i
    }
}        

---

24ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func middleNode(_ head: ListNode?) -> ListNode?
    {
        var x = head
        var arr:[ListNode] = []
        while x != nil {
            arr.append(x!)
            x = x?.next
        }        
        var index = arr.count/2        
        return arr[index]
    }
}