C++ 计算数组中元素重复的次数
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Count how many times elements in an array are repeated
提问by Mahmoud Nabil
The program I'm trying to write allows me to enter 10 numbers and it should get tell me Number X is repeated X times and so on.
我正在尝试编写的程序允许我输入 10 个数字，它应该告诉我数字 X 重复了 X 次等等。
I've been trying this but the problem is I get the result as follows:
我一直在尝试这个，但问题是我得到的结果如下：
For example...{1,1,1,1,4,6,4,7,4}
例如...{1,1,1,1,4,6,4,7,4}
The number 1 is repeated 4 times
The number 1 is repeated 3 times
The number 1 is repeated 2 times
The number 1 is repeated 1 times
The number 4 is repeated 3 times
The number 6 is repeated 1 times
The number 4 is repeated 2 times
The number 7 is repeated 1 times
The number 4 is repeated 1 times
数字 1 重复了 4 次
数字 1 重复了 3 次
数字 1 重复了 2 次
数字 1 重复 1 次
数字 4 重复了 3 次
数字 6 重复 1 次
数字 4 重复了 2 次
数字 7 重复 1 次
数字 4 重复 1 次
The problem is that it checks the next number with the following numbers without skipping it, or without knowing it has written it before
问题是它检查下一个数字与以下数字而不跳过它，或者不知道它之前已经写过
#include <iostream>
#include <string>
using namespace std;
int main() {
int x[10];
for (int i=0;i<10;i++) {
cin>>x[i];
}
for (int i=0;i<9;i++) {
int count=1;
for (int j=i+1;j<10;j++) {
if (x[i]==x[j]) count++;
}
cout<<"The number "<<x[i]<<" is repeated "<<count<<" times"<<"\n";
}
}
回答by saadtaame
The problem with your code is that you re-process numbers that you've already processed. So if there is an occurrence of 1at position 0 and another occurrence of 1at position 5, then you will process the 1at position 5 again when you get there in the loop.
您的代码的问题在于您重新处理了已经处理过的数字。因此，如果1在位置 0 处发生了 ，而1在位置 5 处又发生了一次，那么1当您到达循环中的位置时，您将再次处理位置 5。
So you need a way to decide if a number has been processed already or not. An easy way is to add a second array (initially all values are set to 0) and whenever you process a number you mark all positions where that element occurs. Now before processing an element you check if it's been processed already and do nothing if that's the case.
所以你需要一种方法来决定一个数字是否已经被处理过。一个简单的方法是添加第二个数组（最初所有值都设置为 0），每当您处理一个数字时，您就标记该元素出现的所有位置。现在在处理一个元素之前，你检查它是否已经被处理过，如果是这样的话，什么都不做。
Also, try to indent your code properly :)
另外，请尝试正确缩进您的代码:)
C++ Code:
C++代码：
int main( void ) {
const int N = 10;
int A[N];
for(int i = 0; i < N; i++)
cin >> A[i];
int seen[N];
for(int i = 0; i < N; i++)
seen[i] = 0;
for(int i = 0; i < N; i++) {
if(seen[i] == 0) {
int count = 0;
for(int j = i; j < N; j++)
if(A[j] == A[i]) {
count += 1;
seen[j] = 1;
}
cout << A[i] << " occurs " << count << " times" << endl;
}
}
return 0;
}
回答by andand
Here's a fairly simple implementation using std::map.
这是一个相当简单的实现，使用std::map.
#include <map>
#include <vector>
#include <cstdlib>
#include <iostream>
std::map<int, unsigned int> counter(const std::vector<int>& vals) {
std::map<int, unsigned int> rv;
for (auto val = vals.begin(); val != vals.end(); ++val) {
rv[*val]++;
}
return rv;
}
void display(const std::map<int, unsigned int>& counts) {
for (auto count = counts.begin(); count != counts.end(); ++count) {
std::cout << "Value " << count->first << " has count "
<< count->second << std::endl;
}
}
int main(int argc, char** argv) {
std::vector<int> mem = {1, 1, 1, 1, 4, 6, 4, 7, 4};
display(counter(mem));
return 0;
}
Output:
输出：
Value 1 has count 4
Value 4 has count 3
Value 6 has count 1
Value 7 has count 1
Compiled using the C++14 standard, but it should also work with C++11. Get rid of the vector initializer and use of autoand it should work with C++98.
使用 C++14 标准编译，但它也应该适用于 C++11。摆脱向量初始值设定项和使用，auto它应该适用于 C++98。
回答by Shramana Roy
The most effective way I have recently come across with this...
我最近遇到的最有效的方法...
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int array[10]={1,1,1,1,4,6,4,7,4};
int a[100];
memset(a,0,sizeof(a));
for(int i=0; i<sizeof(array)/sizeof(array[0]); i++)
{
a[array[i]]++;
}
for(int i=1; i<sizeof(a)/sizeof(a[0]); i++)
{
if(a[i]>0)
{
cout<<"The number "<<i<<"is repeated "<<a[i]<<" times"<<"\n";
}
}
OUTPUT:
输出：
The number 1 is repeated 4 times
The number 4 is repeated 3 times
The number 6 is repeated 1 times
The number 7 is repeated 1 times
回答by Abraham Hernandez
Pretty simple using map!
使用非常简单map！
See the Repl.it
参见Repl.it
#include <iostream>
#include <map>
int main()
{
int foo[]{1,1,1,1,4,6,4,7,4};
std::map<int, int> bar;
for (auto const &f : foo)
bar[f]++;
for (auto const &b : bar)
std::cout << "The number " << b.first
<< "is repeated " << b.second
<< "times\n";
}
Expected output:
预期输出：
The number 1 is repeated 4 times
The number 4 is repeated 3 times
The number 6 is repeated 1 times
The number 7 is repeated 1 times
回答by chandresh
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cout<<"enter length of array:"<<endl;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cout<<"enter element:";
cin>>arr[i];
}
sort(arr,arr+n);
/*this is for sort the array so we can find maximum element form user input and
using this element we make one array of that size
*/
int m=arr[n-1];
m++;
int a[m];
for(int i=0;i<m;i++)
{
a[i]=0;
}
for(int i=0;i<n;i++)
{
a[arr[i]]++;
}
cout<<endl;
for(int i=0;i<m;i++)
{
if(a[i]>0)
cout<<i<<"is repeat:"<<a[i]<<"time"<<endl;
}
}
output is like this:
输出是这样的：
enter length of array:
输入数组长度：
6
6
enter element:6
输入元素：6
enter element:5
输入元素：5
enter element:5
输入元素：5
enter element:6
输入元素：6
enter element:2
输入元素：2
enter element:3
输入元素：3
2is repeat:1time
2是重复：1次
3is repeat:1time
3是重复：1次
5is repeat:2time
5是重复：2次
6is repeat:2time
6 次重复：2 次
