C++实现查找中位数的O(N)算法和Kmin算法

本文实例讲述了C++实现查找中位数的O(N)算法和Kmin算法，分享给大家供大家参考。具体方法如下：

利用快速排序的partition操作来完成O(N)时间内的中位数的查找算法如下：

#include <iostream>
#include <cassert>
#include <algorithm>
#include <iterator>

using namespace std;

int array[] = {1, 2, 10, 8, 9, 7, 5};
const int size = sizeof array / sizeof *array;

int partition(int *array, int left, int right)
{
 if (array == NULL)
 return -1;

 int pos = right;
 right--;
 while (left <= right)
 {
 while (left < pos && array[left] <= array[pos])
  left++;

 while (right >= 0 && array[right] > array[pos])
  right--;

 if (left >= right)
  break;

 swap(array[left], array[right]);
 }
 swap(array[left], array[pos]);

 return left;
}

int getMidIndex(int *array, int size)
{
 if (array == NULL || size <= 0)
 return -1;

 int left = 0;
 int right = size - 1;
 int midPos = right >> 1;
 int index = -1;

 while (index != midPos)
 {
 index = partition(array, left, right);

 if (index < midPos)
 {
  left = index + 1;
 }
 else if (index > midPos)
 {
  right = index - 1;
 }
 else
 {
  break;
 }
 }

 assert(index == midPos);
 return array[index];
}

void main()
{
 int value = getMidIndex(array, size);

 cout << "value: " << value << endl;
}

寻找kmin算法如下：

int findKMin(int *array, int size, int k)
{
 if (array == NULL || size <= 0)
 return -1;

 int left = 0;
 int right = size - 1;
 int index = -1;

 while (index != k)
 {
 index = partition(array, left, right);

 if (index < k)
 {
  left = index + 1;
 }
 else if (index > k)
 {
  right = index - 1;
 }
 else
 {
  break;
 }
 }

 assert(index == k);
 return array[index];
}

希望本文所述对大家C++程序算法设计的学习有所帮助。