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分布式与一致性协议之ZAB协议(八)

东方小不点 2024-05-12 阅读 31

 一.判断该链表是否有环(简单):

1.题目

2.思路:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
bool hasCycle(struct ListNode *head) {
    struct ListNode* slow = head,*fast = head;
    while(fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;
        if(fast == slow)
        {
            return true;
        }
    }
    return false;
}

3.扩展问题:

(1) 为什么一定会相遇?有没有可能错过,永远追不上?

 (2) 快指针一次走3步,走四步,...n步,能追上吗?

简单分析一下快指针走三步的情况:

(3)上述是我们理论猜想出来的结果,但是实际真的存在永远追不上的情况吗?

分析:

 重新总结:

二.环形链表2

1.题目

2.思路1: 

我们对指针运动进行数学分析最终得到了头结点到入环节点的距离L与环形链表的关系。

分析代数式,我们得到启发:

注意:

3.参考代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *detectCycle(struct ListNode *head) {
    struct ListNode* slow = head,*fast = head;
    while(fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;
        if(fast == slow)
        {
            struct ListNode* meet = fast;
            while(head != meet)//指针相交时的位置就是入环位置
            {
                head = head->next;
                meet = meet->next;
            }
            return meet;
        }
    }
    return NULL;
}

4.思路二

创建一个新指针newhead,指向meet的next指针,再将meet的next置为NULL。这样就变成了我们所熟悉的相交链表问题。大家可以看我的这篇题解http://t.csdnimg.cn/edNbx

参考代码: 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
//相交链表,如果相交返回相交节点,否则,返回NULL
 struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
    int n1 = 1;
    int n2 = 1;
   struct ListNode* curA = headA;
   struct ListNode* curB = headB;
    while(curA->next)
    {
        curA = curA->next;
        ++n1;
    }
    while(curB->next)
    {
        curB = curB->next;
        ++n2;
    }
    if(curA != curB)
    {//尾节点不相等就不相交
        return NULL;
    }
    //长的先走差距步,再同时走,第一个相等的就是交点
    int num = abs(n1 - n2);
    struct ListNode* LongList = headA;
    struct ListNode* SortList = headB;
    if(n1 < n2)
    {
        LongList = headB;
        SortList = headA;
    }
    while(num--)
    {
        LongList = LongList->next;
    }
    while(LongList != SortList)
    {
        LongList = LongList->next;
        SortList = SortList->next;
    }
    return LongList;
}


struct ListNode *detectCycle(struct ListNode *head) {
    struct ListNode* slow = head,*fast = head;
    while(fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;
        if(fast == slow)
        {
            struct ListNode* meet = fast;
            struct ListNode* newhead = meet->next;
            meet->next = NULL;
            return getIntersectionNode(newhead,head);
        }
    }
    return NULL;
}

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