Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
452 views
in Technique[技术] by (71.8m points)

algorithm - Stability of quicksort partitioning approach

Does the following Quicksort partitioning algorithm result in a stable sort (i.e. does it maintain the relative position of elements with equal values):

  partition(A,p,r)
  {
     x=A[r];
     i=p-1;
     for j=p to r-1
       if(A[j]<=x)
          i++;
          exchange(A[i],A[j])

       exchang(A[i+1],A[r]); 
     return i+1;
   }
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

There is one case in which your partitioning algorithm will make a swap that will change the order of equal values. Here's an image that helps demonstrate how your in-place partitioning algorithm works:

Partition Algorithm

We march through each value with the j index, and if the value we see is less than the partition value, we append it to the light-gray subarray by swapping it with the element that is immediately to the right of the light-gray subarray. The light-gray subarray contains all the elements that are <= the partition value. Now let's look at, say, stage (c) and consider the case in which three 9's are in the beginning of the white zone, followed by a 1. That is, we are about to check whether the 9's are <= the partition value. We look at the first 9 and see that it is not <= 4, so we leave it in place, and march j forward. We look at the next 9 and see that it is not <= 4, so we also leave it in place, and march j forward. We also leave the third 9 in place. Now we look at the 1 and see that it is less than the partition, so we swap it with the first 9. Then to finish the algorithm, we swap the partition value with the value at i+1, which is the second 9. Now we have completed the partition algorithm, and the 9 that was originally third is now first.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...