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）

 

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode FindFirstCommonNode(ListNode phead1, ListNode phead2) {
        
        int head1_length = getlength(phead1);
        int head2_length = getlength(phead2);
        
        if(head1_length>=head2_length)
        {
            int chazhi = head1_length-head2_length;
            for(int i=0;i<chazhi;i++)
            {
                phead1 = phead1.next;
            }
            
        }
        else if(head1_length<head2_length)
        {
            int chazhi = head2_length-head1_length;
            for(int i=0;i<chazhi;i++)
            {
                phead2 = phead2.next;
            }
            
        }
        while(phead1!=phead2)
        {
            phead1 = phead1.next;
            phead2 = phead2.next;
        }
        
        return phead1;
        
 
    }
    
    public static int getlength(ListNode l1)
    {
        int length=0;
        
        while(l1!=null)
        {
            length++;
            l1=l1.next;
        }
        return length;
    }
}