141 lines
4.8 KiB
Markdown
141 lines
4.8 KiB
Markdown
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---
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title: Freecodecamp Algorithm Merge Sort Guide
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localeTitle: Freecodecamp算法合并排序指南
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---
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大多数现代语言都有内置的sort()函数,可以自动对输入数组或列表进行排序。你有没有想过排序功能如何在内部实际工作?了解常见的排序算法及其实现是编码访谈中最重要的部分。在本系列文章中,我们将介绍几种重要的排序算法。它们是如何实现的,时间和空间的复杂性等。我们的第一篇文章是Merge Sort。
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要了解Merge Sort,有关[递归](http://programmers.stackexchange.com/questions/25052/in-plain-english-what-is-recursion)的基本知识是先决条件。 Merge Sort基于Divide and Conquer原则。对N个整数数组进行排序的整个过程可归纳为三个步骤 -
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* 将阵列分成两半。
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* 使用相同的重复算法对左半部分和右半部分进行排序。
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* 合并分类的一半。
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使用Merge排序的最大优点是[时间复杂度](https://www.youtube.com/watch?v=V42FBiohc6c&list=PL2_aWCzGMAwI9HK8YPVBjElbLbI3ufctn)仅为n \* log(n)以对整个Array进行排序。它比冒泡排序或插入排序的n ^ 2运行时间要好很多。
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在编写代码之前,让我们在图表的帮助下理解合并排序的工作原理。
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![合并排序](//discourse-user-assets.s3.amazonaws.com/original/2X/4/4712ef1a5d856dbb4af393fcc08a820a38787395.png)
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* 最初我们有一个包含6个未排序整数的数组Arr(5,8,3,9,1,2)
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* 我们将阵列分成两半Arr1 =(5,8,3)和Arr2 =(9,1,2)。
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* 再次,我们将它们分成两半:Arr3 =(5,8)和Arr4 =(3)和Arr5 =(9,1)和Arr6 =(2)
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* 再次,我们将它们分成两半:Arr7 =(5),Arr8 =(8),Arr9 =(9),Arr10 =(1)和Arr6 =(2)
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* 我们现在将比较这些子数组中的元素以合并它们。
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## 履行
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### C ++实现
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```
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void merge(int array[], int left, int mid, int right)
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{
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int i, j, k;
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// Size of left sublist
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int size_left = mid - left + 1;
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// Size of right sublist
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int size_right = right - mid;
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/* create temp arrays */
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int Left[size_left], Right[size_right];
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/* Copy data to temp arrays L[] and R[] */
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for(i = 0; i < size_left; i++)
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{
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Left[i] = array[left+i];
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}
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for(j = 0; j < size_right; j++)
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{
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Right[j] = array[mid+1+j];
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}
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// Merge the temp arrays back into arr[left..right]
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i = 0; // Initial index of left subarray
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j = 0; // Initial index of right subarray
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k = left; // Initial index of merged subarray
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while (i < size_left && j < size_right)
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{
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if (Left[i] <= Right[j])
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{
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array[k] = Left[i];
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i++;
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}
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else
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{
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array[k] = Right[j];
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j++;
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}
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k++;
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}
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// Copy the remaining elements of Left[]
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while (i < size_left)
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{
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array[k] = Left[i];
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i++;
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k++;
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}
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// Copy the rest elements of R[]
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while (j < size_right)
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{
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array[k] = Right[j];
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j++;
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k++;
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}
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}
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void mergeSort(int array[], int left, int right)
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{
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if(left < right)
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{
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int mid = (left+right)/2;
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// Sort first and second halves
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mergeSort(array, left, mid);
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mergeSort(array, mid+1, right);
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// Finally merge them
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merge(array, left, mid, right);
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}
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}
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/CYVc/1)
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### Javascript实现
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我们在JavaScript中编写MergeSort:
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```
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function mergeSort (arr) {
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if (arr.length < 2) return arr;
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var mid = Math.floor(arr.length /2);
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var subLeft = mergeSort(arr.slice(0,mid));
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var subRight = mergeSort(arr.slice(mid));
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return merge(subLeft, subRight);
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}
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```
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首先,我们检查数组的长度。如果它是1那么我们只返回数组。这将是我们的基本情况。否则,我们将找出中间值并将数组分成两半。我们现在将对MergeSort函数的递归调用对两个部分进行排序。
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```
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function merge (a,b) {
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var result = [];
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while (a.length >0 && b.length >0)
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result.push(a[0] < b[0]? a.shift() : b.shift());
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return result.concat(a.length? a : b);
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}
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```
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当我们合并两个半时,我们将结果存储在辅助数组中。我们将左数组的起始元素与右数组的起始元素进行比较。哪个较小的将被推入结果数组中,我们将使用\[shift()运算符从相应的数组中删除它。如果我们最终得到左或右数组中的值,我们只需在结果的末尾连接它。这是排序结果:
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```
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var test = [5,6,7,3,1,3,15];
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console.log(mergeSort(test));
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>> [1, 3, 3, 5, 6, 7, 15]
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```
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![:rocket:](https://forum.freecodecamp.com/images/emoji/emoji_one/rocket.png?v=3 ":火箭:") [运行代码](https://repl.it/CYVd)
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如果您仍然无法理解MergeSort, [视频说明](https://www.youtube.com/watch?v=TzeBrDU-JaY)将使其更加清晰。
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