由于存储变量的方式,合并排序实现很费劲

问题描述

我有一些C ++和Python的经验,但是肯定我一直在为此苦苦挣扎。我有合并排序的概念,但是我不了解Python如何存储值。为了将这个怪物分成较小的部分,我只关注[11,26,9]的排序。

更具体地说,在下面的输出中,此行之后会发生什么: 1 2 2循环之前1 1 2(i j k)? Python执行哪个循环到达附加输出的最后一行?我的主要困惑是[11]的结果介于9到26之间。是哪个循环引起的?

OUTPUT:
[11,9] left;
[15,17,77] right;
[11,9,15,77] myList after dividing by 2;
[11] left;
[26,9] right;
[11,9] myList after dividing by 2;
[26] left;
[9] right;
[26,9] myList at zero;
1 length of right;
1 length of left;
0 0 0 (i j k) before if loop;
[9,9] myList at else;
[26] left before 2nd while loop;
[9] right before 2nd while loop;
0 1 1  (i j k) before 2nd while loop;
[9,26] myList 2nd while loop i;
[11] left;
[9,26] right;
[11,9] myList at zero;
2 length of right;
1 length of left;
0 0 0 (i j k) before if loop;
[9,9] myList at else;
[11] left;
[9,26] right;
[9,9] myList at zero;
2 length of right;
1 length of left;
0 1 1 (i j k) before if loop;
[9,9] myList at if;
[11] left before 2nd while loop;
[9,26] right before 2nd while loop;
1 1 2  (i j k) before 2nd while loop;//WHAT HAPPENS HERE NEXT WITH [11]?
[11] left after 2nd while loop j;
[9,26] right after 2nd while loop j;
[9,11,26] myList 2nd while loop j;
import scipy.io as sio,math as m,numpy as np,sympy as sym,scipy as sp

def mergeSort(myList):
    if len(myList) > 1:
        mid = len(myList) // 2
        left = myList[:mid]
        print(left,"left")
        right = myList[mid:]
        print(right,"right")
        print(myList,"myList after dividing by 2")
        mergeSort(left)
        mergeSort(right)
        i,j,k = 0,0
     
        while i < len(left) and j < len(right):
            print(left,"left")
            print(right,"right")
            print(myList,"myList at zero")
            print(len(right),"length of right")
            print(len(left),"length of left")
            print(i,k,"(i j k) before if loop")
            # ---------------------------------------------------------------------    
            if left[i] < right[j]:
                print(myList,"myList at if")
                myList[k] = left[i]
                i += 1
            else:
                myList[k] = right[j]
                print(myList,"myList at else")
                j += 1
           
            k += 1
            # -------------------------------------------------------------------------
        print(left,"left before 2nd while loop")
        print(right,"right before 2nd while loop")
        print(i," (i j k) before 2nd while loop")
        # ------------------------------------------------------------------------
        while i < len(left):
            myList[k] = left[i]
            i += 1
            k += 1
            print(myList,"myList 2nd while loop i")
            
        while j < len(right):
            myList[k] = right[j]
            j += 1
            k += 1
            print(left,"left after 2nd while loop j")
            print(right,"right after 2nd while loop j")
            print(myList,"myList 2nd while loop j")
            
myList = [11,77]
mergeSort(myList)
print(myList)

解决方法

我不确定我是否完全理解您的问题,所以如果我的回答不能满足您的问题,请纠正我。

当调用mergeSort()时,长度为3,因此输入被分为左[11]和右[26,9]。此拆分不会以任何方式影响输入myList [11、26、9]。

mergesort内部调用后,左侧仍然是[11],但是右侧更改为[9,26]。

然后,第一个while循环用9覆盖myList中的第一个值。在第二个迭代中,它用11覆盖myList中的第二个值。左,右都不受影响,因此它们仍然保持在左[11]和右[9,26]。 (i,j,k)现在是(1,1,2),因此第一个while循环结束。

第二个while循环被跳过,因此在那里没有变化。

在第三个循环中,myList的剩余第三个值被右边第二个位置的值26覆盖。

这将结束对mySort(...)的原始调用,并且可以在此调用之外使用修改后的列表。

这满足您的问题吗?

PS:我知道这不是您的目标,但是由于python并不是一种性能良好的语言,因此请尝试使用内置函数或库,因为它们中的大多数都是用C / C ++编译的,因此比自编译要快得多。非授权版本:)