java 相交链表的实现示例

1.题目

相交链表：给你两个单链表的头节点 heada 和 headb ，请你找出并返回两个单链表相交的起始节点。如果两个链表没有交点，返回 null 。相交链表

2.分析

相交链表是y字型，next域相同。
定义两个引用pl和ps，

如果每个链表相交结点前长度相同，一步一步走，直到相同就找到了相交结点。如果长度不一样，首先要长链表先走差值步，然后再一人走一步直到相遇

长度不同：

长度相同：

首先求长度，先假设pl指向heada：

 listnode pl = heada;
        listnode ps = headb;

        int lena = 0;
        int lenb = 0;
        while (pl != null) {
            lena++;
            pl = pl.next;
        }
        //pl==null;
        pl = heada;

        while (ps != null) {
            lenb++;
            ps = ps.next;
        }
        //ps==null;
        ps = headb;

然后根据长度差值的正负判断谁长，将pl指向长的链表：

int len = lena - lenb;//差值步
        if (len < 0) {
            pl = headb;
            ps = heada;
            len = lenb - lena;
        }

然后长的先走长度差值步，最后一人一步走：

 //pl走差值len步
        while (len != 0) {
            pl = pl.next;
            len--;
        }
        //同时走，直到相遇
        while (pl != ps) {
            pl = pl.next;
            ps = ps.next;
        }
        return pl;
        }

3.完整代码

//判断链表相交
    public static listnode getintersectionnode(listnode heada, listnode headb) {
        if (heada == null || headb == null) {
            return null;
        }

        listnode pl = heada;
        listnode ps = headb;

        int lena = 0;
        int lenb = 0;
        while (pl != null) {
            lena++;
            pl = pl.next;
        }
        //pl==null;
        pl = heada;

        while (ps != null) {
            lenb++;
            ps = ps.next;
        }
        //ps==null;
        ps = headb;

        int len = lena - lenb;//差值步
        if (len < 0) {
            pl = headb;
            ps = heada;
            len = lenb - lena;
        }
        //1、pl永远指向最长的链表  ps永远指向最短的链表   2、求到了差值len步

        //pl走差值len步
        while (len != 0) {
            pl = pl.next;
            len--;
        }
        //同时走，直到相遇
        while (pl != ps) {
            pl = pl.next;
            ps = ps.next;
        }
        return pl;
    }

测试：

 public static void main(string[] args) {
        mylinkedlist mylinkedlist = new mylinkedlist();
        mylinkedlist.addlast(12);
        mylinkedlist.addlast(23);
        mylinkedlist.addlast(34);
        mylinkedlist.addlast(45);
        system.out.println("mylinkedlist：");
        mylinkedlist.display();

        mylinkedlist mylinkedlist1 = new mylinkedlist();
        mylinkedlist1.addlast(13);
        mylinkedlist1.addlast(22);
        mylinkedlist1.addlast(30);
        system.out.println("mylinkedlist1：");
        mylinkedlist1.display();
        createcut(mylinkedlist.head, mylinkedlist1.head);
        try {
            listnode ret = getintersectionnode(mylinkedlist.head, mylinkedlist1.head);
            mylinkedlist.display2(ret);
        } catch (nullpointerexception e) {
            e.printstacktrace();
            system.out.println("没有相交结点！");
        }

    }

mylinkedlist mylinkedlist = new mylinkedlist();
        mylinkedlist.addlast(12);
        mylinkedlist.addlast(23);
        mylinkedlist.addlast(34);
        mylinkedlist.addlast(56);
        system.out.println("mylinkedlist:");
        mylinkedlist.display();

        mylinkedlist mylinkedlist1 = new mylinkedlist();
        mylinkedlist1.addlast(12);
        mylinkedlist1.addlast(23);
        mylinkedlist1.addlast(30);
        system.out.println("mylinkedlist1:");
        mylinkedlist1.display();
        //createcut(mylinkedlist.head,mylinkedlist1.head);
        try {
            listnode ret = getintersectionnode(mylinkedlist.head, mylinkedlist1.head);
            system.out.println(ret.val);
        }catch (nullpointerexception e){
            e.printstacktrace();
            system.out.println("不存在相交结点！");
        }

    }

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