c++ - 分而治之数组算法C++

Question

我正在尝试实现一个函数，该函数将查看数组的每个元素并确定该特定元素是否大于一个 INT 且小于另一个 INT。例如:

Return true if Arr[5] is >i && < u

我将此作为基本算法，它可以工作，但我想通过使用“分而治之”的方法来创建更有效的代码段，但是我在使用递归使其计数和所有示例时遇到问题我只见过处理一个比较点，而不是两个。谁能解释一下情况。 (http://en.wikipedia.org/wiki/Divide_and_conquer_algorithm)

我的原始代码(线性):

int SimpleSort(int length) 
{ 
    int A[] = {25,5,20,10,50}; 
    int i = 25; //Less Than int 
    u = 2; //Greater Than 
    for(int Count = 0; Count < length; Count++) //Counter 
    { 
        if (A[Count] <= i && A[Count] >= u) //Checker 
            return true; 
    } return false;
}

到目前为止我所获取的示例代码(在处理各种事情并使用不同的示例代码多个小时后没有运气:

int A[] = {5,10,25,30,50,100,200,500,1000,2000};
int i = 10; //Less Than
int u = 5;  //Greater Than


int min = 1;
int max = length;
int mid = (min+max)/2;

if (i < A[mid] && u > A[mid])
{
    min = mid + 1;

}
else
{
    max = mid - 1;
}
Until i <= A1[mid] && u >= A1[mid])

如果这个问题不清楚，抱歉，请询问您是否需要我详细说明。

Answer 1

假设您的输入 vector 总是排序，我认为这样的东西可能适合您。这是我能想到的最简单的形式，性能为 O(log n):

bool inRange(int lval, int uval, int ar[], size_t n)
{
    if (0 == n)
        return false;

    size_t mid = n/2;
    if (ar[mid] >= std::min(lval,uval))
    {
        if (ar[mid] <= std::max(lval,uval))
            return true;
        return inRange(lval, uval, ar, mid);
    }
    return inRange(lval, uval, ar+mid+1, n-mid-1);
}

这使用隐含范围差异；即它始终使用两个值中的较低者作为下限，并使用两个值中的较高者作为上限。如果您的使用要求将 lval 和 uval 的输入值视为福音，那么 any 会调用 lval > uval 应该返回 false(因为这是不可能的)，您可以删除 std::min() 和 std::max() 扩展。无论哪种情况，您都可以通过创建外部前端加载器并预先检查 lval 和 uval 的顺序来进一步提高性能:(a) 如果满足以下条件，则立即返回 false需要绝对排序并且lval > uval，或者(b) 如果需要范围差异，则以适当的顺序预先确定lval 和uval。下面探讨了这两种外包装的示例:

// search for any ar[i] such that (lval <= ar[i] <= uval)
//  assumes ar[] is sorted, and (lval <= uval).
bool inRange_(int lval, int uval, int ar[], size_t n)
{
    if (0 == n)
        return false;

    size_t mid = n/2;
    if (ar[mid] >= lval)
    {
        if (ar[mid] <= uval)
            return true;
        return inRange_(lval, uval, ar, mid);
    }
    return inRange_(lval, uval, ar+mid+1, n-mid-1);
}

// use lval and uval as an hard range of [lval,uval].
//  i.e. short-circuit the impossible case of lower-bound
//  being greater than upper-bound.
bool inRangeAbs(int lval, int uval, int ar[], size_t n)
{
    if (lval > uval)
        return false;
    return inRange_(lval, uval, ar, n);
}

// use lval and uval as un-ordered limits. i.e always use either
// [lval,uval] or [uval,lval], depending on their values.
bool inRange(int lval, int uval, int ar[], size_t n)
{
    return inRange_(std::min(lval,uval), std::max(lval,uval), ar, n);
}

我已将我认为您想要的内容保留为 inRange。下面列出了为覆盖主要情况和边缘情况而执行的单元测试以及结果输出。

#include <iostream>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <iterator>

int main(int argc, char *argv[])
{
    int A[] = {5,10,25,30,50,100,200,500,1000,2000};
    size_t ALen = sizeof(A)/sizeof(A[0]);
    srand((unsigned int)time(NULL));

    // inner boundary tests (should all answer true)
    cout << inRange(5, 25, A, ALen) << endl;
    cout << inRange(1800, 2000, A, ALen) << endl;

    // limit tests (should all answer true)
    cout << inRange(0, 5, A, ALen) << endl;
    cout << inRange(2000, 3000, A, ALen) << endl;

    // midrange tests. (should all answer true)
    cout << inRange(26, 31, A, ALen) << endl;
    cout << inRange(99, 201, A, ALen) << endl;
    cout << inRange(6, 10, A, ALen) << endl;
    cout << inRange(501, 1500, A, ALen) << endl;

    // identity tests. (should all answer true)
    cout << inRange(5, 5, A, ALen) << endl;
    cout << inRange(25, 25, A, ALen) << endl;
    cout << inRange(100, 100, A, ALen) << endl;
    cout << inRange(1000, 1000, A, ALen) << endl;

    // test single-element top-and-bottom cases
    cout << inRange(0,5,A,1) << endl;
    cout << inRange(5,5,A,1) << endl;

    // oo-range tests (should all answer false)
    cout << inRange(1, 4, A, ALen) << endl;
    cout << inRange(2001, 2500, A, ALen) << endl;
    cout << inRange(1, 1, A, 0) << endl;

    // performance on LARGE arrays.
    const size_t N = 2000000;
    cout << "Building array of " << N << " random values." << endl;
    std::vector<int> bigv;
    generate_n(back_inserter(bigv), N, rand);

    // sort the array
    cout << "Sorting array of " << N << " random values." << endl;
    std::sort(bigv.begin(), bigv.end());

    cout << "Running " << N << " identity searches..." << endl;
    for (int i=1;i<N; i++)
        if (!inRange(bigv[i-1],bigv[i],&bigv[0],N))
        {
            cout << "Error: could not find value in range [" << bigv[i-1] << ',' << bigv[i] << "]" << endl;
            break;
        };
    cout << "Finished" << endl;

    return 0;
}

输出结果:

1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
Sorting array of 2000000 random values.
Running 2000000 identity searches...
Finished