C++ 为什么在弗洛伊德'；循环检测的s算法如果我们取快为头。下一步的解会出错吗？

因此，我解决了leetcode的问题，我们必须返回尾部连接的节点，但当我选择

fast=head->next

时，解决方案出现了错误，而当我选择

fast=head

时，解决方案是正确的。我知道后者是正确的，但我不太明白为什么

示例： 输入：头部=[3,2,0，-4]，位置=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;

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;

正确的代码

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;

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;

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你可能会在这方面找到一些线索。我相信这和数学有关，数学给了你循环的起点。这能回答你的问题吗？