相交的链表

下面的代码尝试解决带环和不带环的链表是否相交以及如果相交，交点的位置两个问题。见编程之美3.6节，代码应当同时解决了后面的两个扩展问题。

Node* commonEnd(Node* a, Node* b) {
    if (!a || !b) return 0;

    Node *p1 = a, *p2 = a;
    do {
        if (p2->next) p2 = p2->next;
        if (p2->next) p2 = p2->next;
        if (p1->next) p1 = p1->next;
    } while (p1 != p2);

    Node *q1 = b, *q2 = b;
    do {
        if (q2 == p1) return p1;
        if (q2->next) q2 = q2->next;
        if (q2 == p1) return p1;
        if (q2->next) q2 = q2->next;
        if (q1->next) q1 = q1->next;
    } while (q1 != q2);

    return 0;
}

int distance(Node* a, Node* b) {
    if (a && b) {
        if (a == b) return 0;

        Node* p1 = a, *p2 = a;
        int d = 0;
        do {
            if (p2->next) p2 = p2->next, ++d;
            if (p2 == b) return d;
            if (p2->next) p2 = p2->next, ++d;
            if (p2 == b) return d;
            if (p1->next) p1 = p1->next;
        } while (p1 != p2);
    }

    return -1;
}

Node* step(Node* a, int n) {
    while (n-- && a) a = a->next;
    return a;
}

Node* intersection(Node* a, Node* b) {
    Node* end = commonEnd(a, b);
    if (!end) return 0;
    int da = distance(a, end), db = distance(b, end);

    if (da > db) a = step(a, da - db);
    else b = step(b, db - da);

    while (a != b) a = a->next, b = b->next;
    return a;
}