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  • 《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: 
    ", 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.
    ");
        else
        {
            printf("Expected result: 
    ");
            for (int i = 0; i < k; ++i)
                printf("%d	", expectedResult[i]);
            printf("
    ");
        }
    
        printf("Result for solution1:
    ");
        int* output = new int[k];
        GetLeastNumbers_Solution1(data, n, output, k);
        if (expectedResult != nullptr)
        {
            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("
    
    ");
    }
    
    // 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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  • 原文地址:https://www.cnblogs.com/CJT-blog/p/10514370.html
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