LeetCode #708. Insert into a Cyclic Sorted List_given a sorted file of 1000 records. to insert a n-优快云博客

本文介绍了一种在已排序的循环链表中插入新值的算法，确保链表保持有序状态。算法首先查找链表的最大和最小值，然后根据新值的位置进行插入。若链表为空，则创建新的单节点循环链表。

题目描述：

Given a node from a cyclic linked list which is sorted in ascending order, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be a reference to any single node in the list, and may not be necessarily the smallest value in the cyclic list.

If there are multiple suitable places for insertion, you may choose any place to insert the new value. After the insertion, the cyclic list should remain sorted.

If the list is empty (i.e., given node is null), you should create a new single cyclic list and return the reference to that single node. Otherwise, you should return the original given node.

The following example may help you understand the problem better:

In the figure above, there is a cyclic sorted list of three elements. You are given a reference to the node with value 3, and we need to insert 2 into the list.

The new node should insert between node 1 and node 3. After the insertion, the list should look like this, and we should still return node 3.

class Solution {
public:
    Node* insert(Node* head, int insertVal) {
        if(head==NULL)
        {
            Node* node=new Node(insertVal);
            node->next=node;
            return node;
        }
        // 需要找到链表中的最大值和最小值，从而确定插入的位置
        int max_val=INT_MIN;
        int min_val=INT_MAX;
        Node* p=head->next;
        while(true)
        {
            max_val=max(max_val,p->val);
            min_val=min(min_val,p->val);
            if(p==head) break;
            p=p->next;
        }
        
        Node* cur=head;
        Node* next=cur->next;
        if(insertVal>=max_val||insertVal<=min_val)
        {
            while(true)
            {
                if(cur->val==max_val&&next->val==min_val)
                {
                    Node* node=new Node(insertVal);
                    cur->next=node;
                    node->next=next;
                    break;
                }
                cur=next;
                next=next->next;
            }
        }
        else
        {
            while(true)
            {
                if(cur->val<=insertVal&&next->val>=insertVal)
                {
                    Node* node=new Node(insertVal);
                    cur->next=node;
                    node->next=next;
                    break;
                }
                cur=next;
                next=next->next;
            }
        }
        return head;
    }
};