LeetCode 剑指 Offer 24. 反转链表(swift)

题目

定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。

示例:

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

限制：

0 <= 节点个数 <= 5000

解题思路

public class ListNode {
    public var val: Int
    public var next: ListNode?
    public init(_ val: Int) {
        self.val = val
        self.next = nil
    }
}
/**
 * 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? {
        var frontNode = head
        var backNode = head?.next
        // 边界问题处理
        if backNode == nil || frontNode == nil {
            return frontNode
        }
        // 初始节点指向nil
        frontNode?.next = nil
        while backNode!.next != nil {
            // 临时缓存下一个节点
            let tempNode = backNode!.next
            // 把backNode的指向反转
            backNode?.next = frontNode
            // 把指针移向反转后的节点
            frontNode = backNode
            // 把back指针移到下一个节点
            backNode = tempNode
        }
        backNode!.next = frontNode
        return backNode
    }
}

let node2 = ListNode(2)
let node3 = ListNode(3)
let node4 = ListNode(4)
node2.next = node3
node3.next = node4
let res = Solution().reverseList(node2)
print("res:\(res?.val)\(res!.next!.val)\(res!.next!.next!.val)\(res!.next!.next!.next)")