Median of 3 partitioning

问题

I found the following code for finding a pivot for quicksort using median of first, last and middle element:

int middle = ( low + high ) / 2;
if( a[ middle ].compareTo( a[ low ] ) < 0 )
    swapReferences( a, low, middle );
if( a[ high ].compareTo( a[ low ] ) < 0 )
    swapReferences( a, low, high );
if( a[ high ].compareTo( a[ middle ] ) < 0 )
    swapReferences( a, middle, high );

// Place pivot at position high - 1
swapReferences( a, middle, high - 1 );
Comparable pivot = a[ high - 1 ];

I want to know after finding the median, why is the swap done with index high-1 instead of high?

回答1:

The reason is that the algorithm does not only find the median, it also sorts the low, middle and high elements. After the three permutations you know that a[middle]<=a[high]. So you need only to partition the elements before high, because a[high] is greater or equal to pivot.

Let's look at an example: low=0, middle=4 and high=8. Your array is like this:

lowerOrEqualToPivot X X X pivot X X X greaterOrEqualToPivot

If you swap middle with high, you need to partition the 8 elements between brackets :

[ lowerOrEqualToPivot X X X greaterOrEqualToPivot X X X ] pivot

If you swap middle with high-1, you need to split only 7 elements:

[ lowerOrEqualToPivot X X X X X X ] pivot greaterOrEqualToPivot

By the way there is a bug in the first line:

int middle = ( low + high ) / 2; //Wrong
int middle = ( low + high ) >>> 1; //Correct

The reason is that if (low + high) is greater than Integer.MAX_VALUE you will have an overflow and middle will be a negative number. The second line will always give you a positive result.

来源：https://stackoverflow.com/questions/13144484/median-of-3-partitioning