zoukankan      html  css  js  c++  java
  • 作业分析:(python)Engineering: Algorithms1

    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)
  • 相关阅读:
    js学习总结----案例之拖拽
    面向对象-数据属性
    Apply和call方法-扩充函数赖以生存的作用域
    JS中的function
    JS数组
    JS需要注意的细节和一些基础知识
    策略模式+简单工厂模式
    多态
    MVC3学习 八 Action和result过滤器及日志处理
    MVC3学习 七 JQuery方式和微软自带的AJAX请求
  • 原文地址:https://www.cnblogs.com/ecoflex/p/9205502.html
Copyright © 2011-2022 走看看