队列操作
给定N个整数，任务是将这些元素插入队列中。此外，给定M个整数，任务是找到队列中每个数字的频率。
例子：
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Input:
N = 8
1 2 3 4 5 2 3 1 (N integers)
M = 5
1 3 2 9 10 (M integers)
Output:
2 2 2 -1 -1
Explanation:
After inserting 1, 2, 3, 4, 5, 2, 3, 1 into the queue, frequency of 1 is 2, 3 is 2, 2 is 2, 9 is -1 and 10 is -1 (since, 9 and 10 is not there in the queue).
Input:
N = 5
5 2 2 4 5 (N integers)
M = 5
2 3 4 5 1 (M integers)
Output:
2 -1 1 2 -1
方法：给定的问题可以通过使用一个简单的队列来解决。思路是将N个整数一个一个的插入到队列中，然后检查M个整数出现的频率。将创建单独的函数来执行这两种操作。请按照以下步骤解决问题。
- 创建一个队列并将所有给定的整数推送到队列中。
- 在将所有元素推入队列后，创建另一个函数来一个一个地计算给定M个元素的频率。
- 使用变量cntFrequency来跟踪当前元素的频率。
- 从队列中弹出所有元素直到它的大小，每次检查队列中的当前元素是否等于我们正在检查频率的元素。
– 如果两者相等，则将 cntFrequency增加 1
——否则，什么都不做。 - 将当前元素推回队列，以便下一个案例队列不会受到干扰。
- 打印找到的每个元素的频率。
下面是上述方法的实现：
Java
// Java Program to implement above approach
import java.io.*;
import java.util.*;
class GFG {
// Driver code
public static void main(String[] args)
{
// Declaring Queue
Queue q = new LinkedList<>();
int N = 8;
int[] a = new int[] { 1, 2, 3, 4, 5, 2, 3, 1 };
int M = 5;
int[] b = new int[] { 1, 3, 2, 9, 10 };
// Invoking object of Geeks class
Geeks obj = new Geeks();
for (int i = 0; i < N; i++) {
// calling insert()
// to add elements in queue
obj.insert(q, a[i]);
}
for (int i = 0; i < M; i++) {
// calling findFrequency()
int f = obj.findFrequency(q, b[i]);
if (f != 0) {
// variable f
// will have final frequency
System.out.print(f + " ");
}
else {
System.out.print("-1"
+ " ");
}
}
}
}
// Helper class Geeks to implement
// insert() and findFrequency()
class Geeks {
// Function to insert
// element into the queue
static void insert(Queue q, int k)
{
// adding N integers into the Queue
q.add(k);
}
// Function to find frequency of an element
// return the frequency of k
static int findFrequency(Queue q, int k)
{
// to count frequency of elements
int cntFrequency = 0;
// storing size of queue in a variable
int size = q.size();
// running loop until size becomes zero
while (size-- != 0) {
// storing and deleting
// first element from queue
int x = q.poll();
// comparing if it's equal to integer K
// that belongs to M
if (x == k) {
// increament count
cntFrequency++;
}
// add element back to queue because
// we also want N integers
q.add(x);
}
// return the count
return cntFrequency;
}
} Python3
# Python Program to implement above approach
import collections
# Helper class Geeks to implement
# insert() and findFrequency()
class Geeks:
# Function to insert
# element into the queue
def insert(self, q, k):
# adding N integers into the Queue
q.append(k)
# Function to find frequency of an element
# return the frequency of k
def findFrequency(self, q, k):
# to count frequency of elements
cntFrequency = 0
# storing size of queue in a variable
size = len(q)
# running loop until size becomes zero
while (size):
size = size - 1
# storing and deleting
# first element from queue
x = q.popleft()
# comparing if it's equal to integer K
# that belongs to M
if (x == k):
# increament count
cntFrequency += 1
# add element back to queue because
# we also want N integers
q.append(x)
# return the count
return cntFrequency
# Declaring Queue
q = collections.deque()
N = 8
a = [1, 2, 3, 4, 5, 2, 3, 1]
M = 5
b = [1, 3, 2, 9, 10]
# Invoking object of Geeks class
obj = Geeks()
for i in range(N):
# calling insert()
# to add elements in queue
obj.insert(q, a[i])
for i in range(M):
# calling findFrequency()
f = obj.findFrequency(q, b[i])
if (f != 0):
# variable f
# will have final frequency
print(f, end=" ")
else:
print("-1", end=" ")
print(" ")
# This code is contributed by gfgking.C#
// C# Program to implement above approach
using System;
using System.Collections.Generic;
public class GFG {
// Helper class Geeks to implement
// insert() and findFrequency()
public class Geeks {
// Function to insert
// element into the queue
public void insert(Queue q, int k)
{
// adding N integers into the Queue
q.Enqueue(k);
}
// Function to find frequency of an element
// return the frequency of k
public int findFrequency(Queue q, int k)
{
// to count frequency of elements
int cntFrequency = 0;
// storing size of queue in a variable
int size = q.Count;
// running loop until size becomes zero
while (size-- != 0) {
// storing and deleting
// first element from queue
int x = q.Peek();
q.Dequeue();
// comparing if it's equal to integer K
// that belongs to M
if (x == k) {
// increament count
cntFrequency++;
}
// add element back to queue because
// we also want N integers
q.Enqueue(x);
}
// return the count
return cntFrequency;
}
}
// Driver code
public static void Main()
{
// Declaring Queue
Queue q = new Queue();
int N = 8;
int[] a = new int[] { 1, 2, 3, 4, 5, 2, 3, 1 };
int M = 5;
int[] b = new int[] { 1, 3, 2, 9, 10 };
// Invoking object of Geeks class
Geeks obj = new Geeks();
for (int i = 0; i < N; i++)
{
// calling insert()
// to add elements in queue
obj.insert(q, a[i]);
}
for (int i = 0; i < M; i++)
{
// calling findFrequency()
int f = obj.findFrequency(q, b[i]);
if (f != 0)
{
// variable f
// will have final frequency
Console.Write(f + " ");
}
else {
Console.Write("-1"
+ " ");
}
}
}
}
// This code is contributed by SURENDRA_GANGWAR. 输出
2 2 2 -1 -1 时间复杂度：O(N*M)
辅助空间： O(N)