160. Intersection of Two Linked Lists

算出两个list的长度，然后让长的那个往前移动长度差别。

这样两个list就是从同样长度的地方开始同时往后移动，如果两个头合在一起，或者到了linkedlist的尾部，就返回结果

    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if(headA == null || headB == null) {
            return null;
        }
        int lenA = listLength(headA);
        int lenB = listLength(headB);
        int diff = 0;
        if(lenA > lenB) {
            diff = lenA - lenB;
            while(diff-- > 0) {
                headA = headA.next;
            }
        } else {
            diff = lenB - lenA;
            while(diff-- > 0) {
                headB = headB.next;
            }
        }
        while(headA != null) {
            if(headA == headB) {
                return headA;
            }
            headA = headA.next;
            headB = headB.next;
        }
        return null;
    }
    
    private int listLength(ListNode head) {
        int res = 0;
        ListNode tempHead = head;
        while(tempHead != null) {
            tempHead = tempHead.next;
            res++;
        }
        return res;
    }