该程序应该向用户询问一定数量的值,然后再次向用户询问这些值,对它们进行排序,然后用户将输入一个值。该程序应该搜索这个值并返回该值的位置或告诉用户它不是他给出的列表的一部分。
我使用 notepad++ 和 pocketc++ 来运行和编译整个程序,然后我将它转移到一个 UNIX 编辑器、putty 和 g++ 中,这就是我遇到问题的地方。当我使用 putty 和 g++ 时,程序编译得很好,但只会返回中间排序的值。例如我想要 3 个数字,1 2 3。然后我问程序 2 在哪里,它返回 2 在第二个位置,但任何其他数字都会给我一个垃圾值。这仅适用于 putty 和 c++,而 pocketc++ 将适用于所有值。
#include <iostream>
#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
int main() {
char answer;
int list,target, first, last;
int *array;
int search(int , int , int , int );
int sort( int a[], int);
int index_of_smallest(const int array[], int first, int target);
void swap(int& v1, int& v2);
int search(int data[], int target, int first, int last);
cout << "How many integers does your list have?\n";
cin >> list;
array = new int[list];
cout << "Please enter your integers: ";
for (int i = 0; i < list; i++)
cin >> array[i];
do {
cout << "\nWhat is the target value?\n";
cin >> target;
sort(array,list);
first = 0;
last = list-1;
int k= search( array, target, first, last);
if (k == -1) cout << "That number is not a part of your list of integers.\n";
else cout << "The location of " << target << " is spot " << k+1 << endl;
cout << "Would you like to search a different number?" << endl;
cin >> answer;
} while (answer !='n' && answer !='N');
return 0;
}
int index_of_smallest(const int array[], int first, int target){
int min= array[first];
int index_min=first;
for ( int i = first + 1; i < target; i++)
if (array[i] < min) {min = array[i];
index_min = i;}
return index_min;
}
void swap(int& v1, int& v2){
int temp;
temp = v1;
v1=v2;
v2=temp;
}
void sort(int a[], int num){
int next_smallest;
for (int i= 0; i < num - 1; i++)
{
next_smallest =index_of_smallest(a, i, num);
swap (a[i],a[next_smallest]);
}
}
int search(int data[], int target, int first, int last) {
int middle;
if (first > last)
return -1;
else {
middle=(first + last)/2;
if (target == data[middle])
return middle;
else if ( target < data[middle])
search(data, target, first, middle-1);
else if (target > data[middle])
search(data, target, middle +1, last);}
}