【链表】两个链表的第一个公共结点

输入两个链表，找出它们的第一个公共结点。

public class Solution {
    
    /**
     * 思路：两个链表相交，存在公共的链表尾，根据链表长度的差值，移动指针，找到第一个相同的节点，即为第一个公共节点
     * @param pHead1
     * @param pHead2
     * @return
     */
    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {

        if (pHead1 == null || pHead2 == null) {
            return null;
        }

        int d = 0;

        ListNode p1 = pHead1;
        ListNode p2 = pHead2;

        int length1 = 0;
        while (pHead1 != null) {
            length1++;
            pHead1 = pHead1.next;
        }

        int length2 = 0;
        while (pHead2 != null) {
            length2++;
            pHead2 = pHead2.next;
        }

        if (length1 > length2) {
            d = length1 - length2;

            while (d > 0) {
                p1 = p1.next;
                d--;
            }
        } else if (length1 < length2) {
            d = length2 - length1;

            while (d > 0) {
                p2 = p2.next;
                d--;
            }
        } else {
            return p1;
        }

        while (p1 != null) {
            if (p1 == p2) {
                break;
            }
            p1 = p1.next;
            p2 = p2.next;
        }
        return p1;
    }
}

class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}