计算两个列表共有但价格不同的商品的数量
原文：https://www.geeksforgeeks.org/count-items-common-lists-different-prices/
给定两个列表，分别包含m和n个项目的list1和list2。 每个项目都与两个字段关联：名称和价格。 问题是要计算两个列表中价格不同的商品。
示例：
Input : list1[] = {{"apple", 60}, {"bread", 20},
{"wheat", 50}, {"oil", 30}}
list2[] = {{"milk", 20}, {"bread", 15},
{"wheat", 40}, {"apple", 60}}
Output : 2
bread and wheat are the two items common to both the
lists but with different prices.
方法 1（朴素方法）：使用两个嵌套循环将list1的每个项目与list2的所有项目进行比较。 如果找到价格不同的匹配项，则增加count。
C++
// C++ implementation to count items common to both
// the lists but with different prices
#include <bits/stdc++.h>
using namespace std;
// details of an item
struct item
{
string name;
int price;
};
// function to count items common to both
// the lists but with different prices
int countItems(item list1[], int m,
item list2[], int n)
{
int count = 0;
// for each item of 'list1' check if it is in 'list2'
// but with a different price
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if ((list1[i].name.compare(list2[j].name) == 0) &&
(list1[i].price != list2[j].price))
count++;
// required count of items
return count;
}
// Driver program to test above
int main()
{
item list1[] = {{"apple", 60}, {"bread", 20},
{"wheat", 50}, {"oil", 30}};
item list2[] = {{"milk", 20}, {"bread", 15},
{"wheat", 40}, {"apple", 60}};
int m = sizeof(list1) / sizeof(list1[0]);
int n = sizeof(list2) / sizeof(list2[0]);
cout << "Count = "
<< countItems(list1, m, list2, n);
return 0;
}
Java
// Java implementation to count items common to both
// the lists but with different prices
class GFG{
// details of an item
static class item
{
String name;
int price;
public item(String name, int price) {
this.name = name;
this.price = price;
}
};
// function to count items common to both
// the lists but with different prices
static int countItems(item list1[], int m,
item list2[], int n)
{
int count = 0;
// for each item of 'list1' check if it is in 'list2'
// but with a different price
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if ((list1[i].name.compareTo(list2[j].name) == 0) &&
(list1[i].price != list2[j].price))
count++;
// required count of items
return count;
}
// Driver code
public static void main(String[] args)
{
item list1[] = {new item("apple", 60), new item("bread", 20),
new item("wheat", 50), new item("oil", 30)};
item list2[] = {new item("milk", 20), new item("bread", 15),
new item("wheat", 40), new item("apple", 60)};
int m = list1.length;
int n = list2.length;
System.out.print("Count = "
+ countItems(list1, m, list2, n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python implementation to
# count items common to both
# the lists but with different
# prices
# function to count items
# common to both
# the lists but with different prices
def countItems(list1, list2):
count = 0
# for each item of 'list1'
# check if it is in 'list2'
# but with a different price
for i in list1:
for j in list2:
if i[0] == j[0] and i[1] != j[1]:
count += 1
# required count of items
return count
# Driver program to test above
list1 = [("apple", 60), ("bread", 20),
("wheat", 50), ("oil", 30)]
list2 = [("milk", 20), ("bread", 15),
("wheat", 40), ("apple", 60)]
print("Count = ", countItems(list1, list2))
# This code is contributed by Ansu Kumari.
C
// C# implementation to count items common to both
// the lists but with different prices
using System;
class GFG{
// details of an item
class item
{
public String name;
public int price;
public item(String name, int price) {
this.name = name;
this.price = price;
}
};
// function to count items common to both
// the lists but with different prices
static int countItems(item []list1, int m,
item []list2, int n)
{
int count = 0;
// for each item of 'list1' check if it is in 'list2'
// but with a different price
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if ((list1[i].name.CompareTo(list2[j].name) == 0) &&
(list1[i].price != list2[j].price))
count++;
// required count of items
return count;
}
// Driver code
public static void Main(String[] args)
{
item []list1 = {new item("apple", 60), new item("bread", 20),
new item("wheat", 50), new item("oil", 30)};
item []list2 = {new item("milk", 20), new item("bread", 15),
new item("wheat", 40), new item("apple", 60)};
int m = list1.Length;
int n = list2.Length;
Console.Write("Count = "
+ countItems(list1, m, list2, n));
}
}
// This code is contributed by PrinciRaj1992
输出：
Count = 2
时间复杂度：O(m * n)。
辅助空间：O(1)。
方法 2（二分搜索）：按项目名称的字母顺序对list2排序。 现在，使用二分搜索技术针对list1的每个项目检查其是否存在于list2中，并从list2获得价格。 如果价格不同，则增加count。
CPP
// C++ implementation to count items common to both
// the lists but with different prices
#include <bits/stdc++.h>
using namespace std;
// details of an item
struct item
{
string name;
int price;
};
// comparator function used for sorting
bool compare(struct item a, struct item b)
{
return (a.name.compare(b.name) <= 0);
}
// function to search 'str' in 'list2[]'. If it exists then
// price associated with 'str' in 'list2[]' is being
// returned else -1 is returned. Here binary serach
// trechnique is being applied for searching
int binary_search(item list2[], int low, int high, string str)
{
while (low <= high)
{
int mid = (low + high) / 2;
// if true the item 'str' is in 'list2'
if (list2[mid].name.compare(str) == 0)
return list2[mid].price;
else if (list2[mid].name.compare(str) < 0)
low = mid + 1;
else
high = mid - 1;
}
// item 'str' is not in 'list2'
return -1;
}
// function to count items common to both
// the lists but with different prices
int countItems(item list1[], int m,
item list2[], int n)
{
// sort 'list2' in alphabetcal order of
// items name
sort(list2, list2 + n, compare);
// initial count
int count = 0;
for (int i = 0; i < m; i++) {
// get the price of item 'list1[i]' from 'list2'
// if item in not present in second list then
// -1 is being obtained
int r = binary_search(list2, 0, n-1, list1[i].name);
// if item is present in list2 with a
// different price
if ((r != -1) && (r != list1[i].price))
count++;
}
// required count of items
return count;
}
// Driver program to test above
int main()
{
item list1[] = {{"apple", 60}, {"bread", 20},
{"wheat", 50}, {"oil", 30}};
item list2[] = {{"milk", 20}, {"bread", 15},
{"wheat", 40}, {"apple", 60}};
int m = sizeof(list1) / sizeof(list1[0]);
int n = sizeof(list2) / sizeof(list2[0]);
cout << "Count = "
<< countItems(list1, m, list2, n);
return 0;
}
输出：
Count = 2
时间复杂度：(m * log2(n))。
辅助空间：O(1)。
为了提高效率，应对元素数量最大的列表排序，并将其用于二分搜索。
方法 3（有效方法）：使用（项目名称，价格）元组作为（键，值）创建哈希表。 在哈希表中插入list1的元素。 现在，对于list2的每个元素，检查它是否是哈希表。 如果存在，则检查其价格是否与哈希表中的值不同。 如果是这样，则增加count。
C++
// C++ implementation to count items common to both
// the lists but with different prices
#include <bits/stdc++.h>
using namespace std;
// details of an item
struct item
{
string name;
int price;
};
// function to count items common to both
// the lists but with different prices
int countItems(item list1[], int m,
item list2[], int n)
{
// 'um' implemented as hash table that contains
// item name as the key and price as the value
// associated with the key
unordered_map<string, int> um;
int count = 0;
// insert elements of 'list1' in 'um'
for (int i = 0; i < m; i++)
um[list1[i].name] = list1[i].price;
// for each element of 'list2' check if it is
// present in 'um' with a different price
// value
for (int i = 0; i < n; i++)
if ((um.find(list2[i].name) != um.end()) &&
(um[list2[i].name] != list2[i].price))
count++;
// required count of items
return count;
}
// Driver program to test above
int main()
{
item list1[] = {{"apple", 60}, {"bread", 20},
{"wheat", 50}, {"oil", 30}};
item list2[] = {{"milk", 20}, {"bread", 15},
{"wheat", 40}, {"apple", 60}};
int m = sizeof(list1) / sizeof(list1[0]);
int n = sizeof(list2) / sizeof(list2[0]);
cout << "Count = "
<< countItems(list1, m, list2, n);
return 0;
}
Java
// Java implementation to count
// items common to both the lists
// but with different prices
import java.util.*;
class GFG{
// details of an item
static class item
{
String name;
int price;
public item(String name, int price)
{
this.name = name;
this.price = price;
}
};
// function to count items common to both
// the lists but with different prices
static int countItems(item list1[], int m,
item list2[], int n)
{
// 'um' implemented as hash table that contains
// item name as the key and price as the value
// associated with the key
HashMap<String,
Integer> um = new HashMap<>();
int count = 0;
// insert elements of 'list1' in 'um'
for (int i = 0; i < m; i++)
um.put(list1[i].name, list1[i].price);
// for each element of 'list2' check if it is
// present in 'um' with a different price
// value
for (int i = 0; i < n; i++)
if ((um.containsKey(list2[i].name)) &&
(um.get(list2[i].name) != list2[i].price))
count++;
// required count of items
return count;
}
// Driver program to test above
public static void main(String[] args)
{
item list1[] = {new item("apple", 60),
new item("bread", 20),
new item("wheat", 50),
new item("oil", 30)};
item list2[] = {new item("milk", 20),
new item("bread", 15),
new item("wheat", 40),
new item("apple", 60)};
int m = list1.length;
int n = list2.length;
System.out.print("Count = " +
countItems(list1, m,
clist2, n));
}
}
// This code is contributed by gauravrajput1
C
// C# implementation to count
// items common to both the lists
// but with different prices
using System;
using System.Collections.Generic;
class GFG{
// Details of an item
public class item
{
public String name;
public int price;
public item(String name,
int price)
{
this.name = name;
this.price = price;
}
};
// Function to count items common to
// both the lists but with different prices
static int countItems(item []list1, int m,
item []list2, int n)
{
// 'um' implemented as hash table
// that contains item name as the
// key and price as the value
// associated with the key
Dictionary<String,
int> um = new Dictionary<String,
int>();
int count = 0;
// Insert elements of 'list1'
// in 'um'
for (int i = 0; i < m; i++)
um.Add(list1[i].name,
list1[i].price);
// For each element of 'list2'
// check if it is present in
// 'um' with a different price
// value
for (int i = 0; i < n; i++)
if ((um.ContainsKey(list2[i].name)) &&
(um[list2[i].name] != list2[i].price))
count++;
// Required count of items
return count;
}
// Driver code
public static void Main(String[] args)
{
item []list1 = {new item("apple", 60),
new item("bread", 20),
new item("wheat", 50),
new item("oil", 30)};
item []list2 = {new item("milk", 20),
new item("bread", 15),
new item("wheat", 40),
new item("apple", 60)};
int m = list1.Length;
int n = list2.Length;
Console.Write("Count = " +
countItems(list1, m,
list2, n));
}
}
// This code is contributed by shikhasingrajput
输出：
Count = 2
时间复杂度：O(m + n)。
辅助空间：O(m)。
为了提高效率，应在哈希表中插入元素数量最少的列表。
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