The first common node of two linked lists (Java)

The first common node of two linked lists (Java)

Title: Enter two linked lists and find their first common node. (Note that because the incoming data is a linked list, the error test data tips are displayed in other ways to ensure that the incoming data is correct.

import java.util.Stack;

/**
 * @desc 两个链表的第一个公共节点
 * @author zhaoliang
 * @date 20200416
 */
public class Main52 {
    //输入两个链表，找出它们的第一个公共结点。
    // （注意因为传入数据是链表，所以错误测试数据的提示是用其他方式显示的，保证传入数据是正确的）
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
//方法一：
    public ListNode FindFirstCommonNode1(ListNode pHead1, ListNode pHead2) {
        Stack<ListNode> stack1 = new Stack<>();

        Stack<ListNode> stack2 =new Stack<>();

    while (pHead1 !=null){
        stack1.add(pHead1);
        pHead1 = pHead1.next;
    }
    while (pHead2 !=null){
        stack2.add(pHead2);
        pHead2 = pHead2.next;
    }
    ListNode ans = null;

    while (!stack1.isEmpty() && !stack2.isEmpty()){

        if (stack1.peek().val == stack2.peek().val){
            ans = stack1.peek();
            stack1.pop();
            stack2.pop();
        }else {
            break;
        }
    }
        return ans;
    }
//方法二：
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
    int len1 = 0;
    int len2 = 0;

    ListNode removeNode1 = pHead1;

    ListNode removeNode2 = pHead2;

    while (removeNode1 != null){
        len1++;
        removeNode1 = removeNode1.next;
    }

    while (removeNode2 != null){
        len2++;
        removeNode2 = removeNode2.next;
    }

    if (len1 > len2){
        for (int i = 1;i<=len1-len2;i++){
            pHead1 = pHead1.next;
        }
    }else if(len2 > len1){
        for (int i = 1;i<=len2-len1;i++){
            pHead2 = pHead2.next;
        }
    }
    ListNode ans =null;
    while (pHead1 != null){
        if (pHead1.val == pHead2.val){
            ans = pHead1;
            break;
        }else {
            pHead1 = pHead1.next;
            pHead2 = pHead2.next;
        }
    }
    return ans;
    }
}