最小的K个数

题目: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1:我们通过快排找到第k个数，然后比他的小的都在左边，比他大的都在右边。

思路2:应对大数据的情况，首先取数组中前k个数字建立大根堆，建立堆之后，从第k+1个元素开始，和堆顶元素进行比较，如果小于堆顶的元素，那么就替换堆顶的元素

#include "stdafx.h"
#include<iostream>
#include<set>
#include<vector>
#include<stdlib.h>
#include<tchar.h>
#include<exception>
using namespace std;

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;
}

int Partition(int data[], int length, int start, int end)
{
    if(data == NULL || length <= 0 || start < 0 || end >= length)
        //throw std::exception("Invalid Parameters");
        exit(1);

    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;
}

//=====================方法1====================== 
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];
}

//====================方法2======================
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);
            }
        }
    }
}

int main()
{
    
    int data[] = {4, 5, 1, 6, 2, 7, 3, 8};
    int length = sizeof(data) / sizeof(int);
    int k  = 4;
    
    printf("The data is:
");
    for(int i = 0 ; i < length ; ++i)
        printf("%d	", data[i]);
    printf("

");
    
    vector<int> vectorData;
    for(int i = 0; i < length ; ++i)
        vectorData.push_back(data[i]);
        
    printf("Result for solution1:
");
    
    int* output = new int[k];
    GetLeastNumbers_Solution1(data, length, output, k);
    
    for(int i = 0 ; i < k ; ++i)
        printf("%d	", output[i]);
    printf("

");

    delete[] output;
    
    printf("Result for solution2:
");
    
    intSet leastNumbers;
    GetLeastNumbers_Solution2(vectorData, leastNumbers, k);
    
    printf("The actual output numbers are:
");
    
    for(setIterator iter = leastNumbers.begin(); iter != leastNumbers.end(); ++iter)
        printf("%d	", *iter);
    printf("

");

    
    return 0;
}