两个长度不一定相等的链表，找出其第一个公共节点。

/*public class ListNode {    int val;    ListNode next = null;    ListNode(int val) {        this.val = val;    }}*/public class Solution {    /*    //第一种，这种代码也只有大牛写的出来    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) { 		ListNode p1=pHead1;        ListNode p2=pHead2;        while(p1!=p2){            p1=(p1==null?pHead2:p1.next);            p2=(p2==null?pHead1:p2.next);        }        return p1;    }*/    //第二种方法    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {        int len1=getListLength(pHead1);        int len2=getListLength(pHead2);        int dis=0;        ListNode longHead=null,shortHead=null;        //先分辨链表的长短        if(len1>len2){            dis=len1-len2;            longHead=pHead1;            shortHead=pHead2;        }else{            dis=len2-len1;            longHead=pHead2;            shortHead=pHead1;        }        //先检测较长链表比短链表多出来的那些节点        while(dis>0){            if(longHead==shortHead){                return longHead;            }            dis--;            longHead=longHead.next;        }        //至此两个链表将会同时到达末尾        while(longHead!=null){            if(longHead==shortHead){                return longHead;            }            longHead=longHead.next;            shortHead=shortHead.next;        }        return longHead;       	    }    //计算列表长度    public int getListLength(ListNode phead){        ListNode head=phead;        int count=0;        while(head!=null){            count++;            head=head.next;        }        return count;    }}