Wrie a function which returns the most frequent number in a
list of integers. Handle the case of more than one number
which meets this criterion.
public static int[] GetFrequency(int[] list)
Answer Posted / rob lange
// I added a size input because I don't know how it would
// work without it. I also don't know what you're going
// to do without the returned size information, but whatever.
int* FindMax( int * list, int size )
{
// Check bad input conditions
if ( size == 0 )
return NULL;
else if ( size == 1 )
return list;
// Turn int pointer list into vector
int* curr = list;
std::vector< int > vList;
while ( size != 0 )
{
vList.push_back( *curr );
curr++;
size--;
}
// Sort
sort( vList.begin(), vList.end() );
vList.push_back( 0 ); // dummy tail node
// Find most frequent number
std::vector<int>::iterator i = vList.begin();
int curMaxLength = 0;
int tmpLength = 0;
std::vector<int> MostFreqNums;
while ( i+1 != vList.end() )
{
if ( *i == *(i+1) )
tmpLength++;
else
{
if ( tmpLength > curMaxLength )
{
MostFreqNums.clear();
MostFreqNums.push_back( *i );
curMaxLength = tmpLength;
}
else if ( tmpLength == curMaxLength )
{
MostFreqNums.push_back( *i );
}
tmpLength = 0;
}
++i;
}
// Convert vector list to int pointer array
int * ret = new int[ MostFreqNums.size() ];
i = MostFreqNums.begin();
for( int j = 0; i != MostFreqNums.end(); ++i, ++j )
{
ret[j] = *i;
}
return ret;
}
| Is This Answer Correct ? | 1 Yes | 3 No |
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