*思路： 先求得兩個連結串列的長度，然後得到長度差diff，再先遍歷長連結串列diff步後，再同時遍歷兩個連結串列並比較物件指標。

 1 /*
 2 public class ListNode {
 3     int val;
 4     ListNode next = null;
 5 
 6     ListNode(int val) {
 7         this.val = val;
 8     }
 9 }*/
10 public class Solution {
11     public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
 
12       if(pHead1==null||pHead2==null) 
13         {return null;}
14       ListNode temp1 = pHead1;
15       ListNode temp2 = pHead2;
16       int length1=0;
17       int length2=0;
18       while(temp1!=null){
19           length1++;
20           temp1 = temp1.next;
21       }
22       while(temp2!=null
 
){
23           length2++;
24           temp2 = temp2.next;
25       }
26       int diff= Math.abs(length1-length2);
27       if(length1>=length2){
28           for(int i=0; i<diff; i++){
29               pHead1 = pHead1.next;
30           }
31       }else{
32           for(int i=0; i<diff; i++){
 
33               pHead2 = pHead2.next;
34           }
35       }
36       while(pHead1!=pHead2){
37           pHead1 = pHead1.next;
38           pHead2 = pHead2.next;
39       }
40 
41       return pHead1;
42       
43     }
44 }