LintCode 599:Insert into a Cyclic Sorted List (链表经典题）

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最新推荐文章于 2022-03-01 20:45:22 发布

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分类专栏： algorithm-design 文章标签： LintCode

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本文介绍了一种在已排序的循环链表中插入新节点的算法，确保链表仍保持有序状态。讨论了多种特殊情况，包括重复元素、单节点链表及节点值大于或等于链表所有节点值的情况，并提供了详细的解决方案。

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Insert into a Cyclic Sorted List
Given a node from a cyclic linked list which has been sorted, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be any single node in the list. Return the inserted new node.

Example
Given a list, and insert a value 4:
3->5->1
Return 5->1->3->4

Notice
3->5->1 is a cyclic list, so 3 is next node of 1.
3->5->1 is same with 5->1->3

Solution:
这题很多情况需要处理，注意下面的input case：

10->1->2->4 10 //duplicate case 1
10->1->2->4->4->10->10 //duplicate case 2
null 1
10 1 //single node case
2->2->2 3 // x > all the nodes, and all the nodes the same
3->5->1->2 7 //x > all the nodes
3->5->2->3 1 //x < all the nodes

注意 > 和 == 的情形最好分开写。

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */


class Solution {
public:
    /*
     * @param node: a list node in the list
     * @param x: An integer
     * @return: the inserted new list node
     */
    ListNode * insert(ListNode * node, int x) {

        ListNode *newNode = new ListNode(x);
        if (!node) {
            newNode->next = newNode;
            return newNode;   
        }
        
        ListNode *original = node;
  
        //single node case      
        if (node->next == original) {
            node->next = newNode;
            newNode->next = node;
            return newNode;
        }
        
        while(1) {
            if (node->next == original) { //2->2->2 3
                newNode->next = original->next;
                original->next = newNode;
                break;
            }
            
            if ((x == node->val) ||
                ((x > node->val) && (x < node->next->val)) ||
                ((node->val > node->next->val) && (x > node->val)) ||   //2->2->2 3
                ((node->val > node->next->val) && (x < node->next->val)) //2->2->2 1
                ) { 
                
                newNode->next = node->next;
                node->next = newNode;
                break;
            }
            
            node = node->next;
        }
        
        return newNode;
    }
};