利用快慢指针获取链表闭环的入口

创建两个指针分别指向头节点，同时创建一个临时空指针，当快慢指针相遇的时候（有闭环），将临时空指针指向头节点，同时继续继续下次循环，当慢指针和临时指针相遇，则此节点就是闭环入口。

/// 快慢指针获取环状链表的入口
    private func testEntrance() {
        let firstNode = Node(item: "1", next: nil)
        let secondNode = Node(item: "2", next: nil)
        let thirdNode = Node(item: "3", next: nil)
        let forthNode = Node(item: "4", next: nil)
        let fifthNode = Node(item: "5", next: nil)
        let sixthNode = Node(item: "6", next: nil)
        let seventhNode = Node(item: "7", next: nil)
        let eightNode = Node(item: "8", next: nil)
        
        firstNode.next = secondNode
        secondNode.next = thirdNode
        thirdNode.next = forthNode
        forthNode.next = fifthNode
        fifthNode.next = sixthNode
        sixthNode.next = seventhNode
        seventhNode.next = eightNode
        
        eightNode.next = fifthNode
        
        let node = getEntrance(firstNode: firstNode)
        print("node.item:\(node.item ?? "")")
    }
    
    private func getEntrance(firstNode: Node?) -> Node {
        var fast = firstNode
        var slow = firstNode
        var temp: Node?
        while fast != nil && fast!.next!.next != nil {
            fast = fast!.next!.next!
            slow = slow!.next!
            if fast! == slow! { // 应该是比较节点是否相同
                if temp == nil {    // 被赋值之后就不再管快慢指针了
                    temp = firstNode
                    continue    // 循环体立刻停止本次循环迭代，重新开始下次循环迭代
                }
            }
            if temp != nil {
                temp = temp!.next!
                if slow! == temp! {
                    break
                }
            }
        }
        return slow!
    }

demo地址：https://github.com/yangguanghei/studyDateStructure