归并排序 Merge Sort [译]

本文共有3854个字,关键词:归并排序排序

给定N个项目的数组,Merge Sort将:
1.将每对单独的元素(默认情况下,已排序)合并为2个元素的排序数组(Merge each pair of individual element (which is by default, sorted) into sorted arrays of 2 elements,

2.将每对2个元素的排序数组合并为4个元素的排序数组, 重复这个过程...,(Merge each pair of sorted arrays of 2 elements into sorted arrays of 4 elements, Repeat the process...,

3.最后一步:合并2个N / 2个元素的排序数组(为了简化本讨论,我们假设N是偶数)来获得N个元素的完全排序数组。(Final step: Merge 2 sorted arrays of N/2 elements (for simplicity of this discussion, we assume that N is even) to obtain a fully sorted array of N elements.

This is just the general idea and we need a few more details before we can discuss the true form of Merge Sort.


我们首先讨论归并排序算法的最重要的子程序:O( N )归并,然后解析这个归并排序算法。
We will dissect this Merge Sort algorithm by first discussing its most important sub-routine: The O(N) merge.

给定两个大小为 N1 和 N2 的排序数组 A 和 B,我们可以在O( N ) 时间内将它们有效地归并成一个大小为 N = N1 + N2的组合排序数组。
Given two sorted array, A and B, of size N1 and N2, we can efficiently merge them into one larger combined sorted array of size N = N1+N2, in O(N) time.

这是通过简单地比较两个阵列的头部并始终取两个中较小的一个来实现的。 但是,这个简单但快速的O( N )合并子例程将需要额外的数组来正确地进行合并。
This is achieved by simply comparing the front of the two arrays and take the smaller of the two at all times. However, this simple but fast O(N) merge sub-routine will need additional array to do this merging correctly. See the next slide.

-----子程序代码实现:

void merge(int a[], int low, int mid, int high) {
  // subarray1 = a[low..mid], subarray2 = a[mid+1..high], both sorted
  int N = high-low+1;
  int b[N];        // discuss: why do we need a temporary array b?
  int left = low, right = mid+1, bIdx = 0;
  while (left <= mid && right <= high)       // the merging
    b[bIdx++] = (a[left] <= a[right]) ? a[left++] : a[right++];
  while (left <= mid) b[bIdx++] = a[left++];       // leftover, if any  ※...它解决了我的疑问
  while (right <= high) b[bIdx++] = a[right++];       // leftover, if any
  for (int k = 0; k < N; k++) a[low+k] = b[k];       // copy back

Before we continue, let's talk about Divide(划分) and Conquer(合并) (abbreviated as D&C), a powerful problem solving paradigm.

Divide and Conquer algorithm solves (certain kind of) problem — like our sorting problem — in the following steps:
    1.Divide step: Divide the large, original problem into smaller sub-problems and recursively solve the smaller sub-problems,
(将较大的原始问题划分为较小的子问题,并递归地解决较小的子问题,)
    2.Conquer step: Combine the results of the smaller sub-problems to produce the result of the larger, original problem.
(结合较小子问题的结果,以产生较大的原始问题的结果。)

Merge Sort is a Divide and Conquer sorting algorithm.  

(归并排序是一种分而治之的排序算法)

The divide step is simple: Divide the current array into two halves (perfectly equal if N is even or one side is slightly greater by one element if N is odd) and then recursively sort the two halves.(划分步骤很简单:将当前数组分成两半(如果N是偶数则完全相等,或者如果N是奇数,则一侧稍微大一个元素)然后递归地对两半进行排序。)

The conquer step is the one that does the most work: Merge the two (sorted) halves to form a sorted array, using the merge sub-routine discussed earlier.(合并步骤则是最有效的:使用前面讨论的合并子例程合并两个(已排序)的一半以形成一个有序数组。)

C++实现:

void mergeSort(int a[], int low, int high) {
  // the array to be sorted is a[low..high]
  if (low < high) {      // base case: low >= high (0 or 1 item)
    int mid = (low+high) / 2;    
    mergeSort(a, low  , mid );   // divide into two halves (分成两半)
    mergeSort(a, mid+1, high);    // then recursively sort them(然后递归地对他们进行排序)
    merge(a, low, mid, high);   // conquer : the merge subroutine (合并子程序)
  }
}

from:
     https://visualgo.net/en/sorting?slide=10

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