160 寻找链表交点

解法一(两链表共同后缀长度相同)

    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (headA == nullptr || headB == nullptr)
            return nullptr;
        int len_a = 1, len_b = 1;
        ListNode *temp = headA;
        while ((temp = temp->next))
            ++len_a;
        temp = headB;
        while ((temp = temp->next))
            ++len_b;
        len_a -= len_b;
        while (len_a != 0)
            if (len_a > 0) {
                headA = headA->next;
                --len_a;
            } else {
                headB = headB->next;
                ++len_a;
            }
        while (headA && headB) {
            if (headA == headB)
                return headA;
            headA = headA->next;
            headB = headB->next;
        }
        return nullptr;
    }

解法二

set不能保存重复元素