[Swift]LeetCode206. 反转链表 | Reverse Linked List

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Reverse a singly linked list.

Example:

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

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

---

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

---

20ms

/**
 * 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 reverseList(_ head: ListNode?) -> ListNode? {
        if head == nil || head?.next == nil
        {
            return head
        }
        var h = reverseList(head?.next)
        head?.next?.next = head
        head?.next = nil
        return h
    }
}

---

20ms

/**
 * 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 reverseList(_ head: ListNode?) -> ListNode? {
        if head == nil {
            return nil
        }
        
        var pre : ListNode?
        var next : ListNode?
        var cur : ListNode? = head

        while (cur != nil) {
            next = cur?.next;
            cur?.next = pre;
            pre = cur;
            cur = next;
        }
        
        return pre;
    }
}

---

20ms

/**
 * 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 reverseList(_ head: ListNode?) -> ListNode? {
        if head == nil {
            return head
        }
        
        if head?.next == nil {
            return head
        }
        
        guard let r = reverseList(head?.next) else { return nil }
        if r.next == nil {
            r.next = head
        } else {
            head!.next!.next = head
        }
        head!.next = nil
        return r
    }
}

---

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 reverseList(_ head: ListNode?) -> ListNode? {
        if head == nil || head?.next == nil {
            return head
        }
        
        guard let r = reverseList(head?.next) else { return nil }
        if r.next == nil {
            r.next = head
        } else {
            head!.next!.next = head
        }
        head!.next = nil
        return r
    }
}

---

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 reverseList(_ head: ListNode?) -> ListNode? {
        if head == nil {
            return nil
        }
        var root = head
        
        var stack = [ListNode]()
        while root != nil {
            stack.append(root!)
            root = root!.next
        }
        
        root = stack.last!
        var next = root
        for i in stride(from: stack.count - 2, to: -1, by: -1) {
            let node = stack[i]
            node.next = nil
            next?.next = node
            next = node;
        }
        return root
    }
}