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法律研究C++ 中从两个链表中计数乘积等于给定值的配对
给定两个链表，任务是使用链表的整数元素形成配对，使得它们的乘积等于给定的值，例如 k。链表是一系列数据结构，通过链接连接在一起。
输入
vector v_1 = {5, 7, 8, 10, 11},.
vector v_2 = {6, 4, 3, 2, 0} , int k = 20输出
Count of pairs from two linked lists whose product is equal to a given value k are: 2
解释
The pairs which can be formed using the given linked lists are: (5, 6) = 30(not equals to k), (5, 4) = 20(equals to k), (5, 3) = 15(not equals to k), (5, 2) = 10(not equals to k), (5, 0) = 0(not equals to k), (7, 6) = 42(not equals to k), (7, 4) = 28(not equals to k), (7, 3) = 21(not equals to k), (7, 2) = 14(not equals to k), (7, 0) = 0(not equals to k), (8, 6) = 48(not equals to k), (8, 4) = 32(not equals to k), (8, 3) = 24(not equals to k), (8, 2) = 16(not equals to k), (8, 0) = 0(not equals to k), (10, 6) = 60(not equals to k), (10, 4) = 40(not equals to k), (10, 3) = 30(not equals to k), (10, 2) = 20(not equals to k), (10, 0) = 0(not equals to k), (11, 6) = 66(not equals to k), (11, 4) = 44(not equals to k), (11, 3) = 3(not equals to k), (11, 2) = 22(not equals to k), (11, 0) = 0(not equals to k). So, clearly there are 2 pairs which are equal to the given product.
输入
vector v_1 = {2, 3, 5, 6},.
vector v_2 = {6, 4, 3} , int k = 9输出
Count of pairs from two linked lists whose sum is equal to a given value k are: 1
解释
The pairs which can be formed using the given linked lists are: (2, 6) = 12(not equals to k), (2, 4) = 8(not equals to k), (2, 3) = 6(not equals to k), (3, 6) = 18(not equals to k), (3, 4) = 12(not equals to k), (3, 3) = 9(equals to k), (5, 6) = 30(not equals to k), (5, 4) = 20(not equals to k), (5, 3) = 15(not equals to k), (6, 6) = 36(not equals to k), (6, 4) = 24(not equals to k), (6, 3) = 18(not equals to k),. So, clearly there is 1 pair which is equal to the given sum.
下面程序中使用的方案如下
输入 k 的值和整数类型的值到两个向量中，以便我们可以将向量传递给形成链表。
创建一个函数，该函数将使用作为参数传递给函数的向量来创建一个链表。
遍历循环直到向量的长度，并创建类指针对象
ListNode
遍历循环，直到 ptr->next 不等于 NULL，并将 ptr 设置为 ptr->next
在 ptr->next 中设置 vector[i]
返回开始
创建一个函数，该函数将返回与给定乘积匹配的配对的数量。
取一个临时变量 count 并将其设置为 0
创建两个指针对象，即 *first_list 用于第一个链表，*second_list 用于第二个链表。
从第一个列表的起始指针开始循环，直到列表为空。
在循环内部，从第二个列表的起始指针开始另一个循环，直到列表为空。
在循环内部，检查 IF (first_list->data * second_list->data) == k 则将 count 增加 1
返回 count
打印结果。
示例
#include<bits/stdc++.h>
using namespace std;
class ListNode{
public:
int data;
ListNode *next;
ListNode(int data){
this->data = data;
next = NULL;
}
};
ListNode *CreateList(vector v){
ListNode *start = new ListNode(v[0]);
for (int i = 1; i < v.size(); i++){
ListNode *ptr = start;
while (ptr->next != NULL){
ptr = ptr->next;
}
ptr->next = new ListNode(v[i]);
}
return start;
}
int product_pair(ListNode *start_1, ListNode *start_2, int k){
int count = 0;
ListNode *first_list , *second_list;
for (first_list = start_1; first_list != NULL; first_list = first_list->next){
for (second_list = start_2; second_list != NULL; second_list = second_list->next){
if ((first_list->data * second_list->data) == k){
count++;
}
}
}
return count;
}
int main(){
vector<int> v_1 = {5, 7, 8, 10, 11};
ListNode* start_1 = CreateList(v_1);
vector v_2 = {6, 4, 3, 2, 0};
ListNode* start_2 = CreateList(v_2);
int k = 30;
cout<<"Count of pairs from two linked lists whose product is equal to a given value k are: "<<product_pair(start_1, start_2, k);
}输出
如果我们运行以上代码，它将生成以下输出：
Count of pairs from two linked lists whose product is equal to a given value k are: 2
更新于： 2020-11-02
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