使用C ++ STL在线性时间内查找未排序数组的中位数

给定一个具有N个元素的未排序数组arr [] ，任务是找出线性时间复杂度的数组中位数。

例子：

Input: N = 5, arr[] = {4, 1, 2, 6, 5}
Output: 4
Explanation:
Since N = 5, which is odd, therefore the median is the 3rd element in the sorted array.
The 3rd element in the sorted arr[] is 4.
Hence the median is 4.

Input: N = 8, arr[] = {1, 3, 4, 2, 6, 5, 8, 7}
Output: 4.5
Explanation:
Since N = 8, which is even, therefore median is the average of 4th and 5th element in the sorted array.
The 4th and 5th element in the sorted array is 4 and 5 respectively.
Hence the median is (4+5)/2 = 4.5.

方法：想法是在C++ STL中使用nth_element()函数。

1. 如果数组中元素的数量为奇数，请使用nth_element()函数找到第(N / 2)个元素，如下所示，然后索引(N / 2)的值为给定数组的中位数。

   nth_element(arr.begin(), arr.begin() + N/2, arr.end())

2. 否则，使用nth_element()函数找到第(N / 2)个和第((N – 1)/ 2)个元素，如下所示，并找到索引(N / 2)和((N – 1) / 2)是给定数组的中位数。

   nth_element(arr.begin(), arr.begin() + N/2, arr.end())
   nth_element(arr.begin(), arr.begin() + (N – 1)/2, arr.end())

下面是上述方法的实现：

// C++ program for the above approach
  
#include 
using namespace std;
  
// Function for calculating
// the median
double findMedian(vector a,
                  int n)
{
  
    // If size of the arr[] is even
    if (n % 2 == 0) {
  
        // Applying nth_element
        // on n/2th index
        nth_element(a.begin(),
                    a.begin() + n / 2,
                    a.end());
  
        // Applying nth_element
        // on (n-1)/2 th index
        nth_element(a.begin(),
                    a.begin() + (n - 1) / 2,
                    a.end());
  
        // Find the average of value at
        // index N/2 and (N-1)/2
        return (double)(a[(n - 1) / 2]
                        + a[n / 2])
               / 2.0;
    }
  
    // If size of the arr[] is odd
    else {
  
        // Applying nth_element
        // on n/2
        nth_element(a.begin(),
                    a.begin() + n / 2,
                    a.end());
  
        // Value at index (N/2)th
        // is the median
        return (double)a[n / 2];
    }
}
  
// Driver Code
int main()
{
    // Given array arr[]
    vector arr = { 1, 3, 4, 2,
                        7, 5, 8, 6 };
  
    // Function Call
    cout << "Median = "
         << findMedian(arr, arr.size())
         << endl;
    return 0;
}

输出：

Median = 4.5

时间复杂度： O(N)
辅助空间复杂度： O(1)

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