剑指offer——两个链表的公共节点

题目链接：输入两个链表，找出它们的第一个公共结点。

解题思路：

找出2个链表的长度，然后让长的先走两个链表的长度差，然后再一起走（因为2个链表用公共的尾部）

 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
13         int head1_length = getlength(phead1);
14         int head2_length = getlength(phead2);
15
16         if(head1_length>=head2_length)
17         {
18             int chazhi = head1_length-head2_length;
19             for(int i=0;i<chazhi;i++)
20             {
21                 phead1 = phead1.next;
22             }
23
24         }
25         else if(head1_length<head2_length)
26         {
27             int chazhi = head2_length-head1_length;
28             for(int i=0;i<chazhi;i++)
29             {
30                 phead2 = phead2.next;
31             }
32
33         }
34         while(phead1!=phead2)
35         {
36             phead1 = phead1.next;
37             phead2 = phead2.next;
38         }
39
40         return phead1;
41
42
43     }
44
45     public static int getlength(ListNode l1)
46     {
47         int length=0;
48
49         while(l1!=null)
50         {
51             length++;
52             l1=l1.next;
53         }
54         return length;
55     }
56 }

原文地址：https://www.cnblogs.com/wangyufeiaichiyu/p/10872679.html