25

我在网上找到了这段代码：

def merge(left, right):
    result = []
    i ,j = 0, 0
    while i < len(left) and j < len(right):
        if left[i] <= right[j]:
            result.append(left[i])
            i += 1
        else:
            result.append(right[j])
            j += 1
    result += left[i:]
    result += right[j:]
    return result

def mergesort(list):
    if len(list) < 2:
        return list
    middle = len(list) / 2
    left = mergesort(list[:middle])
    right = mergesort(list[middle:])
    return merge(left, right)

当我运行它时，它可以 100% 运行。我只是不明白合并排序是如何工作的，或者递归函数如何能够正确地对左右排序。

4

8 回答 8

61

我相信理解归并排序的关键是理解以下原则——我称之为归并原则：

给定两个从最小到最大排序的单独列表 A 和 B，通过反复比较 A 的最小值和 B 的最小值，删除较小的值，并将其附加到 C 上，构造一个列表 C。当一个列表用完时，追加将另一个列表中的剩余项目按顺序移到 C 上。列表 C 也是一个排序列表。

如果你手动解决这个问题几次，你会发现它是正确的。例如：

A = 1, 3
B = 2, 4
C = 
min(min(A), min(B)) = 1

A = 3
B = 2, 4
C = 1
min(min(A), min(B)) = 2

A = 3
B = 4
C = 1, 2
min(min(A), min(B)) = 3

A = 
B = 4
C = 1, 2, 3

现在 A 已用尽，因此用 B 中的剩余值扩展 C：

C = 1, 2, 3, 4

合并原理也很容易证明。A的最小值小于A的所有其他值，B的最小值小于B的所有其他值。如果A的最小值小于B的最小值，那么它也必须小于小于 B 的所有值。因此它小于 A 的所有值和 B 的所有值。

因此，只要您继续将满足这些条件的值附加到 C，您就会得到一个排序列表。这就是merge上面的函数所做的。

现在，根据这个原则，很容易理解一种排序技术，它通过将列表分成更小的列表、对这些列表进行排序、然后将这些排序的列表合并在一起来对列表进行排序。该merge_sort函数只是一个简单的函数，它将一个列表分成两半，对这两个列表进行排序，然后以上述方式将这两个列表合并在一起。

唯一的问题是，因为它是递归的，所以当它对两个子列表进行排序时，它通过将它们传递给自己来做到这一点！如果您在这里难以理解递归，我建议您先研究更简单的问题。但是，如果您已经掌握了递归的基础知识，那么您所要做的就是意识到单项列表已经排序。合并两个单项列表会生成一个已排序的两项列表；合并两个二项列表生成一个排序的四项列表；等等。

于 2012-05-08T17:15:15.593 回答
15

当我偶然发现难以理解算法的工作原理时，我会添加调试输出来检查算法内部到底发生了什么。

这里是带有调试输出的代码。尝试理解递归调用的所有步骤mergesort以及merge输出的作用：

def merge(left, right):
    result = []
    i ,j = 0, 0
    while i < len(left) and j < len(right):
        print('left[i]: {} right[j]: {}'.format(left[i],right[j]))
        if left[i] <= right[j]:
            print('Appending {} to the result'.format(left[i]))           
            result.append(left[i])
            print('result now is {}'.format(result))
            i += 1
            print('i now is {}'.format(i))
        else:
            print('Appending {} to the result'.format(right[j]))
            result.append(right[j])
            print('result now is {}'.format(result))
            j += 1
            print('j now is {}'.format(j))
    print('One of the list is exhausted. Adding the rest of one of the lists.')
    result += left[i:]
    result += right[j:]
    print('result now is {}'.format(result))
    return result

def mergesort(L):
    print('---')
    print('mergesort on {}'.format(L))
    if len(L) < 2:
        print('length is 1: returning the list withouth changing')
        return L
    middle = len(L) / 2
    print('calling mergesort on {}'.format(L[:middle]))
    left = mergesort(L[:middle])
    print('calling mergesort on {}'.format(L[middle:]))
    right = mergesort(L[middle:])
    print('Merging left: {} and right: {}'.format(left,right))
    out = merge(left, right)
    print('exiting mergesort on {}'.format(L))
    print('#---')
    return out


mergesort([6,5,4,3,2,1])

输出：

---
mergesort on [6, 5, 4, 3, 2, 1]
calling mergesort on [6, 5, 4]
---
mergesort on [6, 5, 4]
calling mergesort on [6]
---
mergesort on [6]
length is 1: returning the list withouth changing
calling mergesort on [5, 4]
---
mergesort on [5, 4]
calling mergesort on [5]
---
mergesort on [5]
length is 1: returning the list withouth changing
calling mergesort on [4]
---
mergesort on [4]
length is 1: returning the list withouth changing
Merging left: [5] and right: [4]
left[i]: 5 right[j]: 4
Appending 4 to the result
result now is [4]
j now is 1
One of the list is exhausted. Adding the rest of one of the lists.
result now is [4, 5]
exiting mergesort on [5, 4]
#---
Merging left: [6] and right: [4, 5]
left[i]: 6 right[j]: 4
Appending 4 to the result
result now is [4]
j now is 1
left[i]: 6 right[j]: 5
Appending 5 to the result
result now is [4, 5]
j now is 2
One of the list is exhausted. Adding the rest of one of the lists.
result now is [4, 5, 6]
exiting mergesort on [6, 5, 4]
#---
calling mergesort on [3, 2, 1]
---
mergesort on [3, 2, 1]
calling mergesort on [3]
---
mergesort on [3]
length is 1: returning the list withouth changing
calling mergesort on [2, 1]
---
mergesort on [2, 1]
calling mergesort on [2]
---
mergesort on [2]
length is 1: returning the list withouth changing
calling mergesort on [1]
---
mergesort on [1]
length is 1: returning the list withouth changing
Merging left: [2] and right: [1]
left[i]: 2 right[j]: 1
Appending 1 to the result
result now is [1]
j now is 1
One of the list is exhausted. Adding the rest of one of the lists.
result now is [1, 2]
exiting mergesort on [2, 1]
#---
Merging left: [3] and right: [1, 2]
left[i]: 3 right[j]: 1
Appending 1 to the result
result now is [1]
j now is 1
left[i]: 3 right[j]: 2
Appending 2 to the result
result now is [1, 2]
j now is 2
One of the list is exhausted. Adding the rest of one of the lists.
result now is [1, 2, 3]
exiting mergesort on [3, 2, 1]
#---
Merging left: [4, 5, 6] and right: [1, 2, 3]
left[i]: 4 right[j]: 1
Appending 1 to the result
result now is [1]
j now is 1
left[i]: 4 right[j]: 2
Appending 2 to the result
result now is [1, 2]
j now is 2
left[i]: 4 right[j]: 3
Appending 3 to the result
result now is [1, 2, 3]
j now is 3
One of the list is exhausted. Adding the rest of one of the lists.
result now is [1, 2, 3, 4, 5, 6]
exiting mergesort on [6, 5, 4, 3, 2, 1]
#---
于 2012-05-08T17:10:48.783 回答
4

有几种方法可以帮助自己理解这一点：

在调试器中单步执行代码并观察会发生什么。或者，在纸上一步一步（用一个非常小的例子）并观察会发生什么。

（我个人觉得在纸上做这种事情更有指导意义）

list[:middle]从概念上讲，它是这样工作的：输入列表通过减半（例如前半部分）不断被切成越来越小的部分。每一半都一次又一次地减半，直到它的长度小于 2。即，直到它什么都不是或单个元素。然后通过合并例程将这些单独的部分重新组合在一起，方法是将 2 个子列表附加或交错到result列表中，因此您得到一个排序列表。因为必须对 2 个子列表进行排序，所以追加/交错是一种快速 ( O(n) ) 操作。

这（在我看来）的关键不是合并例程，一旦您了解它的输入将始终被排序，这一点就很明显了。“技巧”（我使用引号，因为它不是技巧，它是计算机科学:-)）是为了保证合并的输入是排序的，你必须继续递归，直到你得到一个必须排序的列表，这就是为什么你一直递归调用mergesort直到列表长度小于 2 个元素。

当您第一次遇到递归和扩展合并排序时，它们可能并不明显。您可能想查阅一本好的算法书籍（例如， DPV可在线合法且免费地获得），但您可以通过逐步阅读您拥有的代码来获得很长的路要走。如果您真的想参与其中，Stanford/Coursera算法课程将很快再次运行，他详细介绍了合并排序。

如果你真的想理解它，请阅读那本书参考的第 2 章，然后扔掉上面的代码并从头开始重新编写。严重地。

于 2012-05-08T16:31:50.157 回答
4

归并排序一直是我最喜欢的算法之一。

你从短的排序序列开始，并继续按顺序将它们合并成更大的排序序列。很简单。

递归部分意味着您正在向后工作 - 从整个序列开始并对两半进行排序。每一半也被拆分，直到序列中只有零或一个元素时排序变得微不足道。正如我在最初的描述中所说，随着递归函数返回排序序列变得更大。

于 2012-05-08T16:41:29.860 回答
2

一张图值一千字，一部动画值一万字。

查看以下来自Wikipedia的动画，该动画将帮助您可视化合并排序算法的实际工作原理。

合并排序

详细的动画，对好奇的排序过程中的每个步骤进行解释。

各种排序算法的另一个有趣的动画。

于 2017-03-17T14:44:11.580 回答
0

基本上你得到你的列表，然后你拆分它然后排序它，但是你递归地应用这个方法，所以你最终会再次拆分它，直到你有一个可以轻松排序的简单集合，然后合并所有简单的解决方案得到一个完全排序的数组。

于 2012-05-08T16:31:55.897 回答
0

您可以在这里很好地了解合并排序的工作原理：

http://www.ee.ryerson.ca/~courses/coe428/sorting/mergesort.html

我希望它有所帮助。

于 2014-10-13T19:53:29.133 回答
0

正如Wikipedia文章所解释的，有许多有价值的方法可以完成归并排序。完成合并的方式还取决于要合并的事物的集合，某些集合启用该集合可以使用的某些工具。

我不会用 Python 来回答这个问题，只是因为我不会写；但是，总体而言，参与“合并排序”算法似乎确实是问题的核心。帮助我的资源是 KITE 关于算法的相当过时的网页（由教授编写），仅仅是因为内容的作者消除了上下文有意义的标识符。

我的答案来自这个资源。

请记住，合并排序算法的工作原理是将提供的集合分开，然后将每个单独的部分再次组合在一起，在重新构建集合时将这些部分相互比较。

这里是“代码”（查看最后的 Java“小提琴”）：

public class MergeSort {

/**
 * @param a     the array to divide
 * @param low   the low INDEX of the array
 * @param high  the high INDEX of the array
 */
public void divide (int[] a, int low, int high, String hilo) {


    /* The if statement, here, determines whether the array has at least two elements (more than one element). The
     * "low" and "high" variables are derived from the bounds of the array "a". So, at the first call, this if 
     * statement will evaluate to true; however, as we continue to divide the array and derive our bounds from the 
     * continually divided array, our bounds will become smaller until we can no longer divide our array (the array 
     * has one element). At this point, the "low" (beginning) and "high" (end) will be the same. And further calls 
     * to the method will immediately return. 
     * 
     * Upon return of control, the call stack is traversed, upward, and the subsequent calls to merge are made as each 
     * merge-eligible call to divide() resolves
     */
    if (low < high) {
        String source = hilo;
        // We now know that we can further divide our array into two equal parts, so we continue to prepare for the division 
        // of the array. REMEMBER, as we progress in the divide function, we are dealing with indexes (positions)

        /* Though the next statement is simple arithmetic, understanding the logic of the statement is integral. Remember, 
         * at this juncture, we know that the array has more than one element; therefore, we want to find the middle of the 
         * array so that we can continue to "divide and conquer" the remaining elements. When two elements are left, the
         * result of the evaluation will be "1". And the element in the first position [0] will be taken as one array and the
         * element at the remaining position [1] will be taken as another, separate array.
         */
        int middle = (low + high) / 2;

        divide(a, low, middle, "low");
        divide(a, middle + 1, high, "high");


        /* Remember, this is only called by those recursive iterations where the if statement evaluated to true. 
         * The call to merge() is only resolved after program control has been handed back to the calling method. 
         */
        merge(a, low, middle, high, source);
    }
}


public void merge (int a[], int low, int middle, int high, String source) {
// Merge, here, is not driven by tiny, "instantiated" sub-arrays. Rather, merge is driven by the indexes of the 
// values in the starting array, itself. Remember, we are organizing the array, itself, and are (obviously
// using the values contained within it. These indexes, as you will see, are all we need to complete the sort.  

    /* Using the respective indexes, we figure out how many elements are contained in each half. In this 
     * implementation, we will always have a half as the only way that merge can be called is if two
     * or more elements of the array are in question. We also create to "temporary" arrays for the 
     * storage of the larger array's elements so we can "play" with them and not propogate our 
     * changes until we are done. 
     */
    int first_half_element_no       = middle - low + 1;
    int second_half_element_no      = high - middle;
    int[] first_half                = new int[first_half_element_no];
    int[] second_half               = new int[second_half_element_no];

    // Here, we extract the elements. 
    for (int i = 0; i < first_half_element_no; i++) {  
        first_half[i] = a[low + i]; 
    }

    for (int i = 0; i < second_half_element_no; i++) {  
        second_half[i] = a[middle + i + 1]; // extract the elements from a
    }

    int current_first_half_index = 0;
    int current_second_half_index = 0;
    int k = low;


    while (current_first_half_index < first_half_element_no || current_second_half_index < second_half_element_no) {

        if (current_first_half_index >= first_half_element_no) {
            a[k++] = second_half[current_second_half_index++];
            continue;
        }

        if (current_second_half_index >= second_half_element_no) {
            a[k++] = first_half[current_first_half_index++];
            continue;
        }

        if (first_half[current_first_half_index] < second_half[current_second_half_index]) {
            a[k++] = first_half[current_first_half_index++];
        } else {
            a[k++] = second_half[current_second_half_index++];
        }
    }
}

我也有一个版本，here，它将打印出有用的信息，并提供对上面发生的事情的更直观的表示。如果有帮助，语法突出显示也更好。

于 2015-08-13T21:34:10.237 回答