Algorithm backup ---- Find the kth largest number(寻找第K大数)

　　Here is a way to find the k-th largest number from an ayyay based on the theory of quick sort with an algorithmic complexity of O(n).

　　First we find a base element from the array randomly, and then reorder the list so that all elements which are greater than the pivot come before the pivot (S_a part) and all elements less than the pivot come after it(S_b part). Now we get two cases:

　　1. Element number of S_a part is smaller than K, So the answer is the (k-|S_a|)-th number in S_b;

　　2. If |S_a|>=K, so return the K-th largest number of S_a.

　　Below is the implementation:

/// <summary>
/// Patition according to quick sort algrithm
/// </summary>
/// <param name="numbers">numbers array to be parted</param>
/// <param name="begin">start subscript</param>
/// <param name="end">end subscript</param>
/// <returns>partition point</returns>
public static int Partition(int[] numbers, int begin, int end)
{
    int key = numbers[begin];
    while(begin < end)
    {
        while(begin < end && key >= numbers[end])
            end--;
        if (begin < end) 
            numbers[begin] = numbers[end];
        while(begin < end && key <= numbers[begin])
            begin++;
        if (begin < end)
            numbers[end] = numbers[begin];
    }
    numbers[begin] = key;    
    return begin;
}

/// <summary>
/// Get the k-th largest number from an unsorted array
/// </summary>
/// <param name="numbers">numbers array</param>
/// <param name="begin">start subscript</param>
/// <param name="end">end subscript</param>
/// <param name="k">Point which number to find</param>
/// <returns>the k-th largest number</returns>
public static int GetNumber(int[] numbers, int begin, int end, int k)
{
    //Find the partition according to quick-sort algorithm.
    int n = Partition(numbers, begin, end);
    
    if (n + 1 == k)
    {
        return numbers[n];
    }
    else if (n + 1 < k)
    {
        return GetNumber(numbers, n + 1, end, k);
    }
    else
    {
        return GetNumber(numbers, begin, n - 1, k);
    }
}

 

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