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

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

同时考虑时间和空间的复杂性。 归并排序: 时间复杂度:O(nlogn), 空间复杂度:O(nlogn)

快速排序: 时间复杂度:O(n²), 空间复杂度:O(n)

现在,他们各自在一个场景中获胜。 但是,使用随机枢轴,您几乎总是可以将快速排序的时间复杂度降低到O(nlogn)。

因此,在许多应用中,快速排序是首选,而不是归并排序。

正如其他人所注意到的,快速排序的最坏情况是O(n²),而归并排序和堆排序则停留在O(nlogn)。然而,在平均情况下,这三个都是O(nlogn);所以它们在大多数情况下是可比较的。

平均而言,快速排序更好的地方在于,内循环意味着将多个值与单个值进行比较,而在其他两个循环中,每次比较时两个项都是不同的。换句话说,Quicksort的读取次数是其他两种算法的一半。在现代cpu上,访问时间在很大程度上决定了性能,因此快速排序最终成为一个很好的首选。

在c/c++领域,当不使用stl容器时,我倾向于使用快速排序,因为它是构建的 进入运行时,而归并排序没有。

所以我相信,在许多情况下,这只是阻力最小的途径。

此外,对于整个数据集不适合工作集的情况,快速排序的性能可以高得多。

快速排序并不比归并排序好。对于O(n²)(很少发生的最坏情况),快速排序可能比归并排序的O(nlogn)慢得多。快速排序的开销更小,所以对于小n和速度较慢的计算机,它会更好。但是今天的计算机是如此之快,以至于合并排序的额外开销可以忽略不计,并且在大多数情况下,非常慢的快速排序的风险远远超过合并排序的微不足道的开销。

此外,归并排序将具有相同键的项按原始顺序保留,这是一个有用的属性。