c++ - 为什么在 Floyd 的循环检测算法中，如果我们将 fast 作为 head.next，那么解决方案就会出错？

Question

所以我在解决 leetcode 的问题，我们必须返回尾部连接的节点，但是当我取时解决方案出错了fast = head->next，当我取fast = head. 我明白后者是如何正确的，但我不太明白为什么？

示例： 输入：head = [3,2,0,-4], pos = 1 输出：tail 连接到节点索引 1

错误代码

ListNode *slow = head, *fast = head->next;
    
    while(fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
        
        if(slow == fast)
            break;
    }
    if(fast == NULL || fast->next == NULL){
        return NULL;
    }
    slow = head;
    while(fast != slow){
        slow = slow->next;
        fast = fast->next;
    }
    return fast;

正确的代码

ListNode *slow = head, *fast = head;
    
    while(fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
        
        if(slow == fast)
            break;
    }
    if(fast == NULL || fast->next == NULL){
        return NULL;
    }
    slow = head;
    while(fast != slow){
        slow = slow->next;
        fast = fast->next;
    }
    return fast;