leetcode234和141题，均使用快慢指针思路，为什么退出条件不一样？

Question

假设链表为[1,2,3,4,5,6,7]
那么按照141题的退出条件
while (fast && fast->next)
最后fast指向7退出循环
使用234题的退出条件
while (fast->next && fast->next->next)
仍然是fast指向7退出循环
但是在实际操作中对141题使用234的条件或者对234使用141的条件均会报错，请问为什么？
第141题代码

 class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode *slow = head, *fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) return true;
        }
        return false;
    }
};

第234题代码

 class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if (!head || !head->next) return true;
        ListNode *slow = head, *fast = head;
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode *last = slow->next, *pre = head;
        while (last->next) {
            ListNode *tmp = last->next;
            last->next = tmp->next;
            tmp->next = slow->next;
            slow->next = tmp;
        }
        while (slow->next) {
            slow = slow->next;
            if (pre->val != slow->val) return false;
            pre = pre->next;
        }
        return true;
    }
};

Answer 1

你没有给出代码，我找到了一个

class Solution110_1 {//设置快慢指针，翻转后一半列表
public:
    ListNode *reverselist(ListNode *head)
    {
        ListNode *pre = NULL;
        ListNode *next = NULL;
        while (head!=NULL)
        {
            next = head->next;
            head->next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
    bool isPalindrome(ListNode* head) {
        if (head == NULL || head->next == NULL)
            return true;
        ListNode *slow = head;
        ListNode *fast = head;
        while (fast->next!=NULL && fast->next->next!=NULL)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        slow->next = reverselist(slow->next);
        slow = slow->next;
        while (slow!=NULL)
        {
            if (slow->val != head->val)
                return false;
            slow = slow->next;
            head = head->next;
        }
        return true;
    }
};

循环体里
slow = slow->next;
fast = fast->next->next;
这里fast还是可以通过两个next找到最后的。
关键看你的循环体怎么写。

你调试下就知道了。