Java带头节点单链表的增删融合以及是否有环

class Link<T extends Comparable<T>> {

    // 指向单链表的第一个节点
    Entry<T> head;

    public Link() {
        this.head = new Entry<>(null, null);
    }

    /**
     * 单链表的头插法
     *
     * @param val
     */
    public void insertHead(T val) {
        Entry<T> newNode = new Entry<>(val, null);
        newNode.next = this.head.next;
        this.head.next = newNode;
    }

    /**
     * 单链表的尾插法
     *
     * @param val
     */
    public void insertTail(T val) {
        Entry<T> temp = this.head;
        Entry<T> newNode = new Entry<>(val, null);
        while (temp.next != null) {
            temp = temp.next;
        }
        temp.next = newNode;
    }

    /**
     * 单链表中删除值为val的节点
     *
     * @param val
     */
    public void remove(T val) {
        Entry<T> frontTemp = this.head;
        Entry<T> temp = frontTemp.next;
        while (temp.data != val) {
            frontTemp = frontTemp.next;
            temp = temp.next;
        }
        frontTemp.next = temp.next;
    }

    /**
     * 逆置单链表
     */
    public void reverse() {
        if (this.head.next == null) {
            return;
        }
        Entry<T> cur = this.head.next.next;
        this.head.next.next = null;
        Entry<T> post = null;
        while (cur != null) {
            post = cur.next;
            cur.next = head.next;
            head.next = cur;
            cur = post;
        }
    }

    /**
     * 获取倒数第K个单链表节点的值
     * 1.遍历一次链表，统计链表节点的个数length
     * 2.length-k
     * 只遍历一次链表，找出数
     *
     * @param k
     */
    public T getLastK(int k) {
        Entry<T> cur1 = this.head.next;
        Entry<T> cur2 = cur1.next;
        for (int i = 1; i < k; i++) {
            cur2 = cur2.next;
            if (cur2 == null) {
                return null;
            }
        }
        while (cur2 != null) {
            cur1 = cur1.next;
            cur2 = cur2.next;
        }
        return cur1.data;
    }

    /**
     * 判断链表是否有环，如果有，返回入环节点的值；否则返回null
     *
     * @return
     */
    public T isLinkCircle() {
        //利用快慢指针检测是否有环
        Entry<T> slow = this.head.next;
        Entry<T> quick = this.head.next;
        while (quick != null && quick.next != null) {
            slow = slow.next;
            quick = quick.next.next;
            if (slow == quick) {
                break;
            }
        }
        if (quick == null) {
            return null;
        } else {
            // fast从第一个节点开始走，slow从快慢指针相交的地方开始走，它们相遇的时候，就是环的入口节点
            quick = this.head.next;
            while (quick != slow) {
                quick = quick.next;
                slow = slow.next;
            }
            return slow.data;
        }
    }

    /**
     * 打印单链表的所有元素的值
     */
    public void show() {
        Entry<T> temp = this.head.next;
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
        System.out.println();
    }


    /**
     * 单链表的节点类型
     *
     * @param <T>
     */
    static class Entry<T> {
        T data; // 链表节点的数据域
        Entry<T> next; // 下一个节点的引用

        public Entry(T data, Entry<T> next) {
            this.data = data;
            this.next = next;
        }

    }

    /**
     * 头节点的类型
     *
     * @param <T>
     */
    private static class HeadEntry<T> extends Entry<T> {
        int cnt; // 用来记录节点的个数

        public HeadEntry(int cnt, T data, Entry<T> next) {
            super(data, next);
            this.cnt = cnt;
        }
    }

    /**
     * 合并两个有序的单链表
     *
     * @param link
     */

    public void merge(Link<T> link) {
        Entry<T> p = this.head;
        Entry<T> p1 = this.head.next;
        Entry<T> p2 = link.head.next;

        // 比较p1和p2节点的值，把值小的节点挂在p的后面
        while (p1 != null && p2 != null) {
            if (p1.data.compareTo(p2.data) >= 0) {
                p.next = p2;
                p2 = p2.next;
            } else {
                p.next = p1;
                p1 = p1.next;
            }
            p = p.next;
        }
        if (p1 != null) { // 链表1还有剩余节点
            p.next = p1;
        }

        if (p2 != null) { // 链表2还有剩余节点
            p.next = p2;
        }
    }
}