单向链表快慢指针实际应用问题

快慢指针

所谓快慢指针：就是利用两个指针移动速度的不同来实现需求，一般设置两个指针，慢指针每次移动一格，快指针每次移动两格。下面分享利用快慢指针解决中间值、链表环路以及环路入口的问题。

中间值问题

利用快慢指针，当快指针移动到尾部的时候，慢指针所在的位置即为链表的中间位置

//实现快慢指针查找中间值
public class FastSlow {

    public static void main(String[] args) {
        //创建结点
        Node<String> first = new Node<String>("aa", null);
        Node<String> second = new Node<String>("bb", null);
        Node<String> third = new Node<String>("cc", null);
        Node<String> fourth = new Node<String>("dd", null);
        Node<String> fifth = new Node<String>("ee", null);
        Node<String> six = new Node<String>("ff", null);
        Node<String> seven = new Node<String>("gg", null);
        Node<String> eight = new Node<String>("hh", null);

        //完成结点之间的指向
        first.next = second;
        second.next = third;
        third.next = fourth;
        fourth.next = fifth;
        fifth.next = six;
        six.next = seven;
        seven.next = eight;

        System.out.println("中间值是："+getMid(first));
    }

    //查找中间值，传入的为头节点
    public static String getMid(Node head){
        //定义两个指针，fast一次移动两步，slow一次移动一步
        Node<String> fast = head;
        Node<String> slow = head;
        //使用两个指针进行遍历，当fast移动到尾部的时候，slow则为中间
        while (fast!=null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow.item;
    }
}

单向链表有环问题

原理：因为快指针始终走比慢指针快，那么如果有环，则一定会相遇，即如果快慢指针指向了同一个节点，证明有环

public class CircleCheck {

    public static void main(String[] args) {
        //创建结点
        Node<String> first = new Node<String>("aa", null);
        Node<String> second = new Node<String>("bb", null);
        Node<String> third = new Node<String>("cc", null);
        Node<String> fourth = new Node<String>("dd", null);
        Node<String> fifth = new Node<String>("ee", null);
        Node<String> six = new Node<String>("ff", null);
        Node<String> seven = new Node<String>("gg", null);
        Node<String> eight = new Node<String>("hh", null);

        //完成结点之间的指向
        first.next = second;
        second.next = third;
        third.next = fourth;
        fourth.next = fifth;
        fifth.next = six;
        six.next = seven;

        //创造环
        seven.next = third;

        System.out.println("链表是否有环："+isCircle(first));
    }

    /**
     * @description 校验链表是否有环
     * @params [head]头节点
     * @returns boolean 返回是否有环
     */
    public static boolean isCircle(Node<String> head){
        //定义快慢指针
        Node<String> fast = head;
        Node<String> slow = head;

        //遍历链表，如果快慢指针指向了同一个节点，那么证明有环
        while (fast!=null && fast.next!=null){
            //变换fast和slow指向
            fast = fast.next.next;
            slow = slow.next;

            if (fast.equals(slow)){
                return true;
            }
        }
        return false;
    }
}

有环链表入口问题

当快慢指针相遇时，我们可以判断到链表中有环，这时重新设定一个新指针指向链表的起点，且步长与慢指针一样为1，则慢指针与“新”指针相遇的地方就是环的入口

public class CircleCheck {

    public static void main(String[] args) {
        //创建结点
        Node<String> first = new Node<String>("aa", null);
        Node<String> second = new Node<String>("bb", null);
        Node<String> third = new Node<String>("cc", null);
        Node<String> fourth = new Node<String>("dd", null);
        Node<String> fifth = new Node<String>("ee", null);
        Node<String> six = new Node<String>("ff", null);
        Node<String> seven = new Node<String>("gg", null);
        Node<String> eight = new Node<String>("hh", null);

        //完成结点之间的指向
        first.next = second;
        second.next = third;
        third.next = fourth;
        fourth.next = fifth;
        fifth.next = six;
        six.next = seven;

        //创造环
        seven.next = third;

        System.out.println("链表环的入口为："+getEntrance(first).item);
    }

    /**
     * @description 找到有环链表的入口
     * @date 2023/4/24 9:49
     * @params [head] 头节点
     * @returns Node 返回入口的值
     */
    public static Node getEntrance(Node<String> head){
        //定义快慢指针
        Node<String> fast = head;
        Node<String> slow = head;
        Node<String> temp = null;
        //遍历链表，先找到环，随后准备一个临时指针指向链表的首节点
        //继续遍历，当临时指针和慢指针相遇的时候，指向的节点就是环的入口
        while (fast!=null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;

            //判断快慢指针是否相遇
            if (fast.equals(slow)){
                temp = head;
                continue;
            }

            //判断慢指针是否和临时指针相遇
            if (temp!=null){
                temp = temp.next;
                if (temp.equals(slow)){
                    break;
                }
            }
        }
        return temp;
    }
}

参考
黑马程序员Java数据结构与java算法全套教程