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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