1. 问题描述
   寻找两个链表相交的结点
2. 解决思路
   ①先求出两个链表的长度
   ②将两链表长度相减, 将两链表中较长的链表走两个链表长度的差值的步数, 这时两链表长度相等.
   ③两链表一步一步走, 比较结点是否相同.
   若相同则返回该节点;
   如果直到两个链表都走到结尾了, 还不相同, 则 两链表不相交.
3. 源代码

public class GetIntersectionNode {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        int length1 = getLength(headA);
        int length2 = getLength(headB);
        ListNode l1 = headA;
        ListNode l2 = headB;
        if (length1 > length2) {
            int steps = length1 - length2;
            for (int i = 0; i < steps; i++) {
                l1 = l1.next;
            }
        } else {
            int steps = length2 - length1;
            for (int i = 0; i < steps; i++) {
                l2 = l2.next;
            }
        }
        while (l1 != null && l2 != null) {
            if (l1 == l2) {
                return l1;
            }
            l1 = l1.next;
            l2 = l2.next;
        }
        return null;
    }
    public int getLength(ListNode head) {
        int length = 0;
        while (head != null) {
            length++;
            head = head.next;
        }
        return length;
    }
}