如何找到数组中所有重复元素的位置？

AhmedO 发表于 Dev

艾哈迈多

我可以对代码进行哪些更改以查找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;
    }

chen

如果不允许使用向量，另一种解决方案是让您的函数接受start和end（size）索引。并在main中执行while循环，以从返回的前一个索引+ 1开始搜索。（请注意，这种技术通常可以与STL算法一起使用，例如std::find）

int findnumber(int Array[], int keyword, int start, int size) {

    for (int y = start; 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, 0, 9);

    if (result >= 0) {
        while (result >= 0) {
            cout << "The number " << my_array[result] << " was found in index of : " << result << endl;
            result = findnumber(my_array, searchinput, result+1, 9);
        }
    } else {
        cout << "The number " << searchinput << " was not found :( " << endl;
    }
}

本文收集自互联网，转载请注明来源。

如有侵权，请联系 [email protected] 删除。