go语言判断单链表中是否存在环

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判断单链表是否存在环是一个经典的快慢指针问题，一个每次走一步的指针，和一个每次走两步的指针，如果链表里有环的话，两个指针最终肯定会相遇，具体实现如下：
单链表的结构

type Node struct {
    value    int
    nextNode *Node
}

是否存在环

func hasCycle(head *Node) bool {
    slow := head
    fast := head
    for fast != nil && fast.nextNode != nil {
        slow = slow.nextNode
        fast = fast.nextNode.nextNode
        if fast == slow {
            return true
        }
    }
    return false
}

创建5个节点，并链起来，最后让第四个节点指向第2个节点，

func main() {
    node1 := new(Node)
    node2 := new(Node)
    node3 := new(Node)
    node4 := new(Node)
    node5 := new(Node)
    node1.value = 1
    node2.value = 2
    node3.value = 3
    node4.value = 4
    node5.value = 5
    node1.nextNode = node2
    node2.nextNode = node3
    node3.nextNode = node4
    node4.nextNode = node2
    //node4.nextNode = node5

    fmt.Println(hasCycle(node1))
}

运行结果：

true

说明存在环

再让第四个节点指向第五个几点，就输出了false，说明不存在环

func main() {
    node1 := new(Node)
    node2 := new(Node)
    node3 := new(Node)
    node4 := new(Node)
    node5 := new(Node)
    node1.value = 1
    node2.value = 2
    node3.value = 3
    node4.value = 4
    node5.value = 5
    node1.nextNode = node2
    node2.nextNode = node3
    node3.nextNode = node4
    //node4.nextNode = node2
    node4.nextNode = node5

    fmt.Println(hasCycle(node1))
}

完整代码

package main

import "fmt"

type Node struct {
    value    int
    nextNode *Node
}

func hasCycle(head *Node) bool {
    slow := head
    fast := head
    for fast != nil && fast.nextNode != nil {
        slow = slow.nextNode
        fast = fast.nextNode.nextNode
        if fast == slow {
            return true
        }
    }
    return false
}

func main() {
    node1 := new(Node)
    node2 := new(Node)
    node3 := new(Node)
    node4 := new(Node)
    node5 := new(Node)
    node1.value = 1
    node2.value = 2
    node3.value = 3
    node4.value = 4
    node5.value = 5
    node1.nextNode = node2
    node2.nextNode = node3
    node3.nextNode = node4
    //node4.nextNode = node2
    node4.nextNode = node5

    fmt.Println(hasCycle(node1))
}

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