最小的K个数

输入n个数，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，则最小的4个数是1,2,3,4.

解法一：O（n）的算法，只有当我们可以修改输入数组时可用

int Partitoin(int* input, int low, int high)
{
    if (input == NULL || low < 0 || high < 0 || low > high)
    {
        return -1;
    }
    int temp = input[low];
    while(low < high)
    {
        while (high>low && input[high] >= temp)
        {
            high--;
        }
        input[low] = input[high];
        while (low<high && input[low] <= temp)
        {
            low++ ;
        }
        input[high] = input[low];
    }
    input[low] = temp ;

    return low ;
}

void GetLeastNumbers(int* input , int length, int k)
{
    if (input == NULL || length < 1 || k < 0 || k >length)
    {
        return;
    }
    int low = 0 ;
    int high = length - 1 ;
    int index = Partitoin(input, low, high);
    while (index != (k - 1))
    {
        if (index < (k - 1))
        {
            low = index + 1;
            index = Partitoin(input, low, high);
        }
        else
        {
            high = index - 1 ;
            index = Partitoin(input, low, high);
        }
    }
    for (int i = 0 ; i < k ; i++)
    {
        cout<<input[i]<<" ";
    }
}

解法二：O（nlogk）的算法，特别适合处理海量数据

typedef  multiset<int , greater<int> > intSet ;
typedef intSet::iterator  setIterator ;
void GetLeastNumbers(vector<int>& data , intSet& leastNumbers, int k)
{
    if (k < 1 || data.size() < k)
    {
        return;
    }

    vector<int>::iterator ite = data.begin();
    for (;ite != data.end() ; ite++)
    {
        if ( (leastNumbers.size()) < k)
        {
            leastNumbers.insert(*ite);
        }
        else  
        {
            setIterator iterGreatest = leastNumbers.begin() ; 
            if (*ite < *iterGreatest)
            {
                leastNumbers.erase(iterGreatest);
                leastNumbers.insert(*ite);
            }
        }
        
    }
}