《剑指offer》第四十题(最小的k个数)

// 面试题40: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。

#include <set>
#include <vector>
#include <iostream>
#include <functional>

using namespace std;


// ====================方法1====================
//使用Partition函数,利用快排思想找到前K个数
/*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;
}

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

    int index = RandomInRange(start, end);//生成随机点
    Swap(&data[index], &data[end]);//放到最后

    int small = start - 1;
    for (index = start; index < end; ++index)//把数列中小于随机点的值放到前面
    {
        if (data[index] < data[end])
        {
            ++small;
            if (small != index)
                Swap(&data[index], &data[small]);
        }
    }

    ++small;
    Swap(&data[small], &data[end]);//然后把随机点放入小于它的值的后面

    return small;
}

/*核心算法*/
void GetLeastNumbers_Solution1(int* input, int n, int* output, int k)
{
    if (input == nullptr || output == nullptr || 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)//可见改变了输入数组
        output[i] = input[i];
}

// ====================方法2====================
typedef multiset<int, greater<int> >            intSet;//greater函数和less相反,这个谓词
typedef multiset<int, greater<int> >::iterator  setIterator;//设置迭代器

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

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

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

        else
        {
            setIterator iterGreatest = leastNumbers.begin();//multiset迭代器

            if (*iter < *(leastNumbers.begin()))
            {
                leastNumbers.erase(iterGreatest);//删除该条目
                leastNumbers.insert(*iter);//插入新的,自动排序
            }
        }
    }
}

// ====================测试代码====================
void Test(const char* testName, int* data, int n, int* expectedResult, int k)
{
    if (testName != nullptr)
        printf("%s begins: \n", testName);

    vector<int> vectorData;
    for (int i = 0; i < n; ++i)
        vectorData.push_back(data[i]);

    if (expectedResult == nullptr)
        printf("The input is invalid, we don't expect any result.\n");
    else
    {
        printf("Expected result: \n");
        for (int i = 0; i < k; ++i)
            printf("%d\t", expectedResult[i]);
        printf("\n");
    }

    printf("Result for solution1:\n");
    int* output = new int[k];
    GetLeastNumbers_Solution1(data, n, output, k);
    if (expectedResult != nullptr)
    {
        for (int i = 0; i < k; ++i)
            printf("%d\t", output[i]);
        printf("\n");
    }

    delete[] output;

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

// k小于数组的长度
void Test1()
{
    int data[] = { 4, 5, 1, 6, 2, 7, 3, 8 };
    int expected[] = { 1, 2, 3, 4 };
    Test("Test1", data, sizeof(data) / sizeof(int), expected, sizeof(expected) / sizeof(int));
}

// k等于数组的长度
void Test2()
{
    int data[] = { 4, 5, 1, 6, 2, 7, 3, 8 };
    int expected[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
    Test("Test2", data, sizeof(data) / sizeof(int), expected, sizeof(expected) / sizeof(int));
}

// k大于数组的长度
void Test3()
{
    int data[] = { 4, 5, 1, 6, 2, 7, 3, 8 };
    int* expected = nullptr;
    Test("Test3", data, sizeof(data) / sizeof(int), expected, 10);
}

// k等于1
void Test4()
{
    int data[] = { 4, 5, 1, 6, 2, 7, 3, 8 };
    int expected[] = { 1 };
    Test("Test4", data, sizeof(data) / sizeof(int), expected, sizeof(expected) / sizeof(int));
}

// k等于0
void Test5()
{
    int data[] = { 4, 5, 1, 6, 2, 7, 3, 8 };
    int* expected = nullptr;
    Test("Test5", data, sizeof(data) / sizeof(int), expected, 0);
}

// 数组中有相同的数字
void Test6()
{
    int data[] = { 4, 5, 1, 6, 2, 7, 2, 8 };
    int expected[] = { 1, 2 };
    Test("Test6", data, sizeof(data) / sizeof(int), expected, sizeof(expected) / sizeof(int));
}

// 输入空指针
void Test7()
{
    int* expected = nullptr;
    Test("Test7", nullptr, 0, expected, 0);
}

int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();
    Test7();
    system("pause");
    return 0;
}

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转载自www.cnblogs.com/CJT-blog/p/10514370.html