链表相交

给你两个单链表的头节点 headA 和 headB ，请你找出并返回两个单链表相交的起始节点。如果两个链表没有交点，返回 null 。

代码如下：

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        // 先求出链表A和链表B的长度
        ListNode curA = headA;
        ListNode curB = headB;
        int lenA = 0,lenB= 0;
        while (curA != null){
            lenA ++;
            curA = curA.next;
        }
        while (curB != null){
            lenB ++;
            curB = curB.next;
        }
        curA = headA;
        curB = headB;
        // 让curA为最长链表的头，lenA为其长度
        if (lenB > lenA){
            // 1 交换lenA和lenB
            int tmplen = lenA;
            lenA = lenB;
            lenB = tmplen;
            // 2 交换curA和curB
            ListNode tmpNode = curA;
            curA = curB;
            curB = tmpNode;
        }
        // 求长度差
        int gap = lenA - lenB;
        // 让curA和curB在同一起点上（末尾位置对齐）
        while (gap -- > 0 ){
            curA = curA.next;
        }
        // 遍历curA和curB，遇到相同直接返回
        while (curA != null){
            if (curA == curB){
                return curA;
            }
            curA = curA.next;
            curB = curB.next;
        }
        return null;
    }

    class ListNode{
        int val;
        ListNode next;

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

        public ListNode() {
        }

        public ListNode(int val, ListNode next) {
            this.val = val;
            this.next = next;
        }
    }
    }