面试题30：最小的k个数

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。

题目分析

剑指Offer（纪念版）P167

代码实现

 

O(n)的算法，只有当我们可以修改输入的数组时可用

void GetLeastNumbers_Solution1(int* input, int n, int* output, int k)
{
    if(input == NULL || output == NULL || k > n || n <= 0 || k <= 0)
        return;

    int start = 0;
    int end = n - 1;
    int index = Partition(input, n, start, end);
    while(index != k - 1)
    {
        if(index > k - 1)
        {
            end = index - 1;
            index = Partition(input, n, start, end);
        }
        else
        {
            start = index + 1;
            index = Partition(input, n, start, end);
        }
    }

    for(int i = 0; i < k; ++i)
        output[i] = input[i];
}



int Partition(int data[], int length, int start, int end)
{
    if(data == NULL || length <= 0 || start < 0 || end >= length)
        throw new std::exception("Invalid Parameters");

    int index = RandomInRange(start, end);
    Swap(&data[index], &data[end]);

    int small = start - 1;
    for(index = start; index < end; ++ index)
    {
        if(data[index] < data[end])
        {
            ++ small;
            if(small != index)
                Swap(&data[index], &data[small]);
        }
    }

    ++ small;
    Swap(&data[small], &data[end]);

    return small;
}

// Random Partition
int RandomInRange(int min, int max)
{
    int random = rand() % (max - min + 1) + min;
    return random;
}

void Swap(int* num1, int* num2)
{
    int temp = *num1;
    *num1 = *num2;
    *num2 = temp;
}

O(nlogk)的算法，特别适合处理海量数据

下面的代码也可以使用priority_queue来实现

typedef multiset<int, greater<int> >            intSet;
typedef multiset<int, greater<int> >::iterator  setIterator;

void GetLeastNumbers_Solution2(const vector<int>& data, intSet& leastNumbers, int k)
{
    leastNumbers.clear();

    if(k < 1 || data.size() < k)
        return;

    vector<int>::const_iterator iter = data.begin();
    for(; iter != data.end(); ++ iter)
    {
        if((leastNumbers.size()) < k)
            leastNumbers.insert(*iter);

        else
        {
            setIterator iterGreatest = leastNumbers.begin();

            if(*iter < *(leastNumbers.begin()))
            {
                leastNumbers.erase(iterGreatest);
                leastNumbers.insert(*iter);
            }
        }
    }
}