作业分析：（python）Engineering: Algorithms1 - SELF PACED Algorithms: Design and Analysis

Programming Question 1

1/1 point (graded)

Download the text file here. (Right click and select "Save As...")

This file contains all of the 100,000 integers between 1 and 100,000 (inclusive) in some order, with no integer repeated.

Your task is to compute the number of inversions in the file given, where the ith row of the file indicates the ith entry of an array.

Because of the large size of this array, you should implement the fast divide-and-conquer algorithm covered in the video lectures.

The numeric answer for the given input file should be typed in the space below.

So if your answer is 1198233847, then just type 1198233847 in the space provided without any spaces or commas or any other punctuation marks. You can make up to 5 attempts.

(We do not require you to submit your code, so feel free to use any programming language you want --- just type the final numeric answer in the following space.)

[TIP: before submitting, first test the correctness of your program on some small test files or your own devising. Then post your best test cases to the discussion forums to help your fellow students!]

遇到过的坑记录一下：

（1）第一个错误：没有用global变量，而是把每次merge sort的return (sorted_list,inversion)，用tuple的形式传回，导致复杂度变成O(n^2)

（2）第二个错误：用了pop函数偷懒，导致运算时间大约变为原来的2倍

（3）第三个错误：导入txt到Python list的时候，用的ls = readlines（file），此时结果会略微偏小，应该写一个循环，如下

for line in file:

    ls.append(int(line))

 

此时运算结果 2407905288，用时大约0.8s左右

 

import time

inverse = 0

def mergesort_nopop(ls):
    global inverse
    if len(ls) <= 1:
        return ls
    else:
        i = len(ls)//2
        ls1 = mergesort_nopop(ls[:i])
        ls2 = mergesort_nopop(ls[i:])
        ls = []
        i = 0
        j = 0
        while True:
            if ls1[i] <= ls2[j]:
                ls.append(ls1[i])
                i += 1
            else:
                ls.append(ls2[j])
                j += 1
                inverse += len(ls1) - i
            if len(ls1) == i:
                ls += ls2[j:]
                break
            if len(ls2) == j:
                ls += ls1[i:]
                break
        return ls


file = open("D:/python doc/algorithm/IntegerArray.txt","r")
# ls = file.readlines()
ls = []
for line in file:
    ls.append(int(line))


start = time.time()
mergesort_nopop(ls)
# mergesort_nopop([1,3,5,2,4,6])
print("count:",inverse)
end = time.time()
print("total time:",end-start)