我在一次面试中被问到这个问题。它们都是O(nlogn),但大多数人使用快速排序而不是归并排序。为什么呢?


当前回答

One of the reason is more philosophical. Quicksort is Top->Down philosophy. With n elements to sort, there are n! possibilities. With 2 partitions of m & n-m which are mutually exclusive, the number of possibilities go down in several orders of magnitude. m! * (n-m)! is smaller by several orders than n! alone. imagine 5! vs 3! *2!. 5! has 10 times more possibilities than 2 partitions of 2 & 3 each . and extrapolate to 1 million factorial vs 900K!*100K! vs. So instead of worrying about establishing any order within a range or a partition,just establish order at a broader level in partitions and reduce the possibilities within a partition. Any order established earlier within a range will be disturbed later if the partitions themselves are not mutually exclusive.

任何自下而上的排序方法,如归并排序或堆排序,就像工人或雇员的方法一样,人们很早就开始在微观层面进行比较。但是,一旦在它们之间发现了一个元素,这个顺序就必然会丢失。这些方法非常稳定和可预测,但要做一定量的额外工作。

Quick Sort is like Managerial approach where one is not initially concerned about any order , only about meeting a broad criterion with No regard for order. Then the partitions are narrowed until you get a sorted set. The real challenge in Quicksort is in finding a partition or criterion in the dark when you know nothing about the elements to sort. That is why we either need to spend some effort to find a median value or pick 1 at random or some arbitrary "Managerial" approach . To find a perfect median can take significant amount of effort and leads to a stupid bottom up approach again. So Quicksort says just a pick a random pivot and hope that it will be somewhere in the middle or do some work to find median of 3 , 5 or something more to find a better median but do not plan to be perfect & don't waste any time in initially ordering. That seems to do well if you are lucky or sometimes degrades to n^2 when you don't get a median but just take a chance. Any way data is random. right. So I agree more with the top ->down logical approach of quicksort & it turns out that the chance it takes about pivot selection & comparisons that it saves earlier seems to work better more times than any meticulous & thorough stable bottom ->up approach like merge sort. But

其他回答

我想补充的是,到目前为止提到的三种算法(归并排序,快速排序和堆排序)只有归并排序是稳定的。也就是说,对于那些具有相同键的值,顺序不会改变。在某些情况下,这是可取的。

但是,说实话,在实际情况下,大多数人只需要良好的平均性能和快速排序…快速=)

所有排序算法都有其起伏。有关排序算法的概述,请参阅维基百科的文章。

在所有条件相同的情况下,我希望大多数人使用最方便的方法,这往往是qsort(3)。除此之外,快速排序在数组上非常快,就像归并排序是列表的常用选择一样。

我想知道的是为什么基数排序和桶排序这么少见。它们是O(n)至少在链表上是这样的它所需要的只是将键转换为序数的方法。(字符串和浮动工作得很好。)

我认为原因与计算机科学的教学方式有关。我甚至不得不向我的讲师演示算法分析,它确实有可能比O(nlog (n))更快地排序。(他证明了比较排序不能比O(nlog (n))快,这是正确的)

在其他新闻中,浮点数可以按整数排序,但之后必须将负数反转。

编辑: 实际上,这里有一种更糟糕的将浮点数作为整数排序的方法:http://www.stereopsis.com/radix.html。注意,不管你实际使用什么排序算法,比特翻转技巧都可以使用……

与归并排序不同,快速排序不使用辅助空间。而归并排序使用辅助空间O(n)。 归并排序的最坏情况时间复杂度是O(nlogn)而快速排序的最坏情况复杂度是O(n²)这发生在数组已经排序的时候。

答案将略微倾向于快速排序w.r.t的变化带来的DualPivotQuickSort的基本值。它在JAVA 7中用于在JAVA .util. arrays中排序

It is proved that for the Dual-Pivot Quicksort the average number of
comparisons is 2*n*ln(n), the average number of swaps is 0.8*n*ln(n),
whereas classical Quicksort algorithm has 2*n*ln(n) and 1*n*ln(n)
respectively. Full mathematical proof see in attached proof.txt
and proof_add.txt files. Theoretical results are also confirmed
by experimental counting of the operations.

您可以在这里找到JAVA7实现- http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/7-b147/java/util/Arrays.java

关于DualPivotQuickSort的进一步精彩阅读- http://permalink.gmane.org/gmane.comp.java.openjdk.core-libs.devel/2628

快速排序具有更好的平均情况复杂度,但在某些应用中它是错误的选择。快速排序容易受到拒绝服务攻击。如果攻击者可以选择要排序的输入,他可以很容易地构造一个时间复杂度为o(n^2)的最坏情况的集合。

归并排序的平均情况复杂性和最坏情况复杂性是相同的,因此不会遇到相同的问题。归并排序的这一特性也使它成为实时系统的最佳选择——确切地说,因为没有导致它运行得非常非常慢的病理情况。

由于这些原因,我更喜欢归并排序,而不是快速排序。