Question

我正在尝试制作一个二进制搜索程序，我的编码在下面，但是当我给出偶数个元素时，它不会给出任何输出，当我给出奇数个元素时，程序运行得很好!!程序首先比较中间值，如果为假，则比较小于或大于中间值

int main()
{
    int n,a[50] ;
    int i, j, temp,counter,searchv,f,l,mid;
    cout<<"Enter no. of elements: ";
    cin>>n;
    f=0;
    l=n-1;
    mid=(f+l)/2;
    cout<<"l= "<<l<<" mid= "<<mid<<"\n";
    cout<<"Enter "<<n <<" values \n";
    for(counter=0;counter<n;counter++)
    {
        cin>>a[counter];
    }

    for(j=0; j<n; j++)
    {
        for (int i=(n-1); i>j ;i--)
        {
            if (a[i]<a[i-1])
            {
                int temp=a[i-1];
                a[i-1]=a[i];
                a[i]=temp;
            }
        }
    }
    cout<<"SORTED ARRAY!!\n";
    for(counter=f;counter<n;counter++)
    {
        cout<<"Value at Element "<<counter <<" is "<<a[counter];

        cout<<endl;
    }
    cout<<"Enter number to search: ";
    cin>>searchv;

    if(a[mid]==searchv)
    {
        cout<<"searched value "<<searchv<<" founded at position "<<mid;
    }
    else if(searchv>a[mid])
    {
        for(counter=l;counter>mid;counter--)
        {
            if(a[counter]==searchv)
                cout<<"searched value "<<searchv<<" founded at position "<<counter;
            break;
        }
    }
    else if(searchv<a[mid])
    {
        for(counter=0;counter<mid;counter++)
        {
            if(a[counter]==searchv)
                cout<<"searched value "<<searchv<<" founded at position "<<counter;
            break;
        }
    }
    else
    {
        cout<<"Value not found\n";
    }
    getch();
    return 0;
}

Answer 1

您的错误与数组是否具有偶数或奇数元素无关。在你的两个线性搜索循环中，例如：

    for (counter = l; counter > mid; counter--) {
        if (a[counter] == searchv)
            cout << "searched value " << searchv <<
                " founded at position " << counter << '\n';
        break;
    }

你break无条件退出循环。这意味着您只测试范围的第一个元素。你想在你找到匹配项时才打破，当然，你应该使用大括号：

    for (counter = l; counter > mid; counter--) {
        if (a[counter] == searchv) {
            cout << "searched value " << searchv <<
                " founded at position " << counter << '\n';
            break;
        }
    }

如果您养成一致地缩进和使用大括号的习惯，这些错误更容易被发现，可能除了最基本的if没有else之外。

通过此修复程序，您的程序将仅报告匹配项，但不报告任何错误，因为lase else在逻辑上永远不会发生，因为前三个if / else涵盖所有可能的案例。（对于整数，a == b，a < b和a > b涵盖所有情况。）

您可以通过保留一个额外的变量来修复此问题，该变量告诉您是否找到了匹配项，或者将搜索重构为单独的函数，无论如何这都是可行的。

Answer 2

根据上面的答案，您可以添加此

而不是

if(a[mid]==searchv)
    {
    cout<<"searched value "<<searchv<<" founded at position "<<mid;
    }
    else if(searchv>a[mid])
    {
    for(counter=l;counter>mid;counter--)
    {
        if(a[counter]==searchv)
    cout<<"searched value "<<searchv<<" founded at position "<<counter;
    break;
    }
    }
    else if(searchv<a[mid])
    {
    for(counter=0;counter<mid;counter++)
    {
    if(a[counter]==searchv)
    cout<<"searched value "<<searchv<<" founded at position "<<counter;
    break;
    }
    }

else
{
     cout<<"Value not found\n";
}

你可以试试这个：

    bool isFound = false;

    for(counter=0;counter<l;counter++)
    {
        if(a[counter]==searchv)
        {
            cout<<"searched value "<<searchv<<" founded at position "<< counter << endl;
            isFound = true;

        }
    }

    if (isFound == false)
    {
        cout << "No result found." << endl;
    }

当我放置偶数个元素时，为什么我的程序不起作用

2 个答案: