题目描述
输入n个整数,找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字,则最小的4个数字是1,2,3,4。
思路一:基于partition函数的算法,时间复杂度可达到O(n)
class Solution
{
vector<int> GetLeastNumbers_Solution( vector<int> input, int k )
{
if ( k <= 0 )
return vector<int>();
if ( input.size() < k )
return vector<int>();
int start = 0, end = input.size()-1;
int kIndex = k - 1;
int index;
vector<int> output;
while ( kIndex != ( index = partition( input, start, end ) ) )
{
if ( kIndex > index )
start = index+1;
else
end = index-1;
}
for ( int i = 0; i < k; i++ )
{
output.push_back( input[i] );
}
return output;
}
//[start, end]
int partition( vector<int> &input, int start, int end )
{
int sentry = start;
start++;
while ( start <= end )
{
if ( input[start] <= input[sentry] )
{
start++;
continue;
}
if ( input[end] > input[sentry] )
{
end--;
continue;
}
swap( &input[start], &input[end] );
start++;
end--;
}
swap( &input[end], &input[sentry] );
return end;
}
void swap( int *a, int *b )
{
int temp = *a;
*a = *b;
*b = temp;
}
}
思路二:借助最大堆时间复杂度可达到O(nlogk)的算法:
class Solution {
public:
vector<int> GetLeastNumbers_Solution( vector<int> input, int k )
{
intSet leastNumbers;
if ( k < 1 || k > input.size() )
return vector<int>();
vector<int>::const_iterator iter = input.begin();
for ( ; iter != input.end(); ++iter )
{
if ( leastNumbers.size() < k )
leastNumbers.insert( *iter );
else
{
setIterator iterGreatest = leastNumbers.begin();
if ( *iter < *(leastNumbers.begin()) )
{
leastNumbers.erase( iterGreatest );
leastNumbers.insert( *iter );
}
}
}
return vector<int>( leastNumbers.begin(), leastNumbers.end() );
}
private:
typedef multiset<int, greater<int> > intSet;
typedef multiset<int, greater<int> >::iterator setIterator;
};
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