While sorting an array for ex: A[5]={1,4,5,3,2} the output must be 1,2,3,4,5 in ascending order. in using the concept of bubble sorting my output is 0,1,2,3,4 what would be the problem in my code
int A[5]={1,5,3,2,4};
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
if(A[j]>A[j+1])
{
int t=A[j];
A[j]=A[j+1];
A[j+1]=t;
}
}
}
for(i=0;i<5;i++)
cout<<A[i];
You need to limit your inner loop to <4:
int A[5]={1,5,3,2,4};
for(int i=0;i<5;i++){
for(int j=0;j<4;j++){
if(A[j]>A[j+1])
{
int t=A[j];
A[j]=A[j+1];
A[j+1]=t;
}
}
}
for(i=0;i<5;i++)
cout<<A[i];
Why not use the STL sort?
#include <algorithm>
std::sort(A, A+5);
A is an array or vector, then one can write A.size(), but for C-style array the preferred way probably is to #define ARRAYSIZE(A) (sizeof(A)/sizeof(A[0])) and then call std::sort( A, A + ARRAYSIZE(A) );
A is a pointer. If you want to do something like this, use an inline template. Also, I'm not sure why this was downvoted. Seems a sensible answer in the general case. Of course, if the question is a part of a homework assignment or something, the STL version is probably out, but we don't know that.A is a pointer then ARRAYSIZE does not work. But, that's changing the problem definition too much, in the problem A was not a pointer :). Also I mentioned .size() can be used whenever possible. BTW, it's not me who downvoted.Perhaps you are printing i instead of A[i] in the printing loop, as in
for( int i = 0; i < N; i++ ) {
cout << i << ","; // by mistake, printing i instead of A[i]
}
Is there a reason you are doing bubble sort, other then trying to learn it? It's one of the slower sorts.
