c++ - 如何找到数组中所有重复元素的位置？

Question

我可以对我的代码应用哪些更改来查找 EX 的每个重复元素的位置:12 或 45代码:

#include < iostream >
using namespace std;

int   findnumber ( int Array[]  ,  int keyword  , int size ){
    for  ( int y=0 ; y<size ; y++ )
        if  ( keyword==Array[y] ) {
            return y;
        }
    return -1; //not found
}

int main  ( ){
    int my_array[]={12,12,5,6,9,45,5,54,45};
    int searchinput;
    cout<<"please select number to search from these (12,12,5,6,9,45,5,54,45) : ";
    cin>>searchinput;
    int result. =  findnumber  (my_array , searchinput , 9);
    if(result>=0){
        cout<<"The number "<<my_array[result]<<" was found in index of : "<<result<<endl;
    }
    else
    {
    cout<<"The number "<<searchinput<<" was not found :( "<<endl;
    } 

Answer 1

解决方案

更新函数以返回索引 vector 而不是单个索引。一种方法是按如下方式定义它:

#include <vector>


vector<int> findnumber(int Array[], int keyword, int size) {
    vector<int> res;
    for (int y = 0; y < size; y++) {
        if (keyword == Array[y]) {
            res.push_back(y);
        }
    }
    return res; 
}

此外，您的函数调用和打印应修改如下:

vector<int> result = findnumber(my_array, searchinput, 9);
if (result.size() > 0) {
    cout << "The number " << searchinput << " was found in the following indices : ";
    for (int i : result){
        cout << i << " ";
    }
    cout << endl;
}
else{
    cout << "The number " << searchinput << " was not found :( " << endl;
}

结果

input: 45
output: The number 45 was found in the following indices : 5 8
input: 12
output: The number 12 was found in the following indices : 0 1
input: 6
output: The number 6 was found in the following indices : 3
input: 7
output: The number 7 was not found :(