单链表（面试题）

链表反转思路：

 1.单链表中有效节点的个数

2.查找单链表中弟第k个节点

3.单链表的反转

实现代码如下：

public class TestLink {

    public static void main(String[] args) {
        //先创建节点
        HeroNode hero1 = new HeroNode(1, "宋江", "及时雨");
        HeroNode hero2 = new HeroNode(2, "卢俊义", "玉麒麟");
        HeroNode hero3 = new HeroNode(3, "吴用", "智多星");
        HeroNode hero4 = new HeroNode(4, "林冲", "豹子头");
        //创建链表
        SingleLinkedList list = new SingleLinkedList();
        list.addByOrder(hero1);
        list.addByOrder(hero4);
        list.addByOrder(hero3);
        list.addByOrder(hero2);
        /*int aa = TestLink.count(list.getHead());
        System.out.println(aa);
        HeroNode hero5 = TestLink.findHeroNode(list.getHead(),2);
        System.out.println(hero5.toString());*/
        TestLink.reverse(list.getHead());
        list.list();
    }
    
    //1单链表中有效节点的个数
    //--遍历得出个数
    public static int count(HeroNode head){
        if(head.next==null){
            return 0;
        }
        int count = 0;
        while(true){
            head = head.next;
            count++;
            if(head.next==null){
                break;
            }
        }
        return count;
    }
    //2.查找单链表中弟第k个节点
    //--先找到总长度
    //--然后算出在第几个 遍历
    public static HeroNode findHeroNode(HeroNode head,int k){
        if(head.next==null){
            return null;
        }
        int size = count(head);
        if(size<k||k<=0){
            return null;
        }
        int index = 0;
        while(true){
            head = head.next;
            if(index==(size-k)){
                break;
            }
            index++;
        }
        return head;
    }
    //3.单链表的反转
    //--创建一个新的头部
    //--循环列表 逐步把遍历到的插入新的头部后面
    //--把原来的头部 指向新的队列
    public static void reverse(HeroNode head){
        if(head.next==null){
            System.out.println("链表为空");
            return;
        }
        HeroNode rever = new HeroNode(0, "", "");
        HeroNode next = null;
        HeroNode cru = head.next;
        while(cru!=null){
            next = cru.next;
            cru.next = rever.next;
            rever.next=cru;
            cru=next;
        }
        head.next = rever.next;
    }
    

}