2807. Insert Greatest Common Divisors in Linked List-优快云博客

给定一个链表，每个节点包含一个整数值。在每对相邻节点间插入它们最大公约数的值。当没有相邻节点时，返回链表。文章提供了两种方法实现此功能：一种是迭代，另一种是递归。

Given the head of a linked list head, in which each node contains an integer value.

Between every pair of adjacent nodes, insert a new node with a value equal to the greatest common divisor of them.

Return the linked list after insertion.

The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.

Example 1:

Input: head = [18,6,10,3]
Output: [18,6,6,2,10,1,3]
Explanation: The 1st diagram denotes the initial linked list and the 2nd diagram denotes the linked list after inserting the new nodes (nodes in blue are the inserted nodes).
- We insert the greatest common divisor of 18 and 6 = 6 between the 1st and the 2nd nodes.
- We insert the greatest common divisor of 6 and 10 = 2 between the 2nd and the 3rd nodes.
- We insert the greatest common divisor of 10 and 3 = 1 between the 3rd and the 4th nodes.
There are no more adjacent nodes, so we return the linked list.

Example 2:

Input: head = [7]
Output: [7]
Explanation: The 1st diagram denotes the initial linked list and the 2nd diagram denotes the linked list after inserting the new nodes.
There are no pairs of adjacent nodes, so we return the initial linked list.

Constraints:

The number of nodes in the list is in the range [1, 5000].
1 <= Node.val <= 1000

法一：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* insertGreatestCommonDivisors(ListNode* head) {
        int count = 1;
        while(head->next == nullptr){
            return head;
        }
        ListNode * node = head;
        while(node -> next != nullptr){
            int a,b;
            if(node -> val >= node -> next -> val){
                a = node -> val;
                b = node -> next -> val;
            }
            else{
                a = node -> next -> val;
                b = node -> val;
            }
            int temp;
            while(b!= 0){
                temp = a;
                a = b;
                b = temp%b;
            }
            node -> next = new ListNode(a,node->next);
            node = node -> next -> next;
        }        
        return head;
    }
};

法二：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* insertGreatestCommonDivisors(ListNode* head) {
        while(head->next == nullptr){
            return head;
        }
        ListNode * node = head;
        while(node -> next != nullptr){
            node -> next = new ListNode(std::__gcd(node->val, node->next->val),node->next);
            node = node -> next -> next;
        }        
        return head;
    }
};