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如何捅破python编程的那层纸之三
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pythonstudy3.md

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#如何捅破python编程的那层纸之三
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# coding:utf-8
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"""
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如何捅破python编程的那层纸之三
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贴吧上有人问
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定义一个函数,名字为sameSums(aList),alist是一个整形list,函数作用是判断能分成两组,使得两组数字的和相等。若可以择返回值是true,若不可以返回值是false。如下例:
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sameSums([4, 7, 6, 3]) --> True //4+6 = 10 and 7 + 3 = 10
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sameSums([3, 3]) --> True
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sameSums([4, 12, 16]) --> True //4+12= 16 and 16
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sameSums([5, 1]) --> False
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这个题目,对初学者来说,有点难度,但稍微有点算法基础,编程思路,就不难。
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前面用贪心算法,只能满足上面4个测试用例的正确解决。
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如何捅破python编程的那层纸之二
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https://github.com/pythonpeixun/article/blob/master/pythonstudy2.md
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但是对[4, 5, 6, 7, 8],[2, 2, 2, 3, 3]这样的list,不能解答出正确的答案。
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这个问题:可以用递归解题,也可以用动态规划解题,黄哥用这样的思路解一下,供初学者参考!
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解题思路:
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1、如果sum(list) 是奇数,就不能将list拆分为2个和相等的list
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2、如果sum(list)之和为偶数,就 sub_sum = sum(list)/2
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2、列出所有可能的子集(子list)
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3、判断子集是不是和 sub_sum相等,如果相等则可以拆分为2个和相等的list,
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否则不可以
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本文由黄哥python培训,黄哥所写
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黄哥python培训试看视频播放地址
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https://github.com/pythonpeixun/article/blob/master/python_shiping.md
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"""
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import itertools
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def get_all_subset(lst):
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"""求list 所有子集合"""
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tmp_lst = []
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length = len(lst)
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for i in xrange(1, length):
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tmp_lst += (set(itertools.combinations(lst, i)))
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return tmp_lst
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def sameSums(int_list):
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"""黄哥python培训 黄哥所写 qq:1465376564
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>>> sameSums([4, 7, 6, 3])
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True
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>>> sameSums([3, 3])
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True
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>>> sameSums([4, 12, 16])
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True
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>>> sameSums([5, 1])
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False
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"""
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sum_of_lsit = sum(int_list)
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sub_sum = sum_of_lsit / 2
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if sum_of_lsit % 2:
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return False
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all_subset = get_all_subset(int_list)
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for item in all_subset:
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if sum(item) == sub_sum:
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return True
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return False
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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lst1 = [3, 9, 10, 30, 8]
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lst2 = [4, 5, 6, 7, 8]
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lst3 = [2, 2, 2, 3, 3]
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print sameSums(lst1)
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print sameSums(lst2)
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print sameSums(lst3)
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[迪艾姆python培训_python编程思路- ]
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(http://v.youku.com/v_show/id_XNTY0MDA5MDMy.html)
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[迪艾姆python培训_python编程思路二 ]
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(http://v.youku.com/v_show/id_XNTY0MDE1NzA0.html)
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[python爬虫之采集搜素引擎联想词 ]
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(http://www.tudou.com/programs/view/SXgshk-sYbw/)
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[广告:如果自学有困难,建议参加迪艾姆python培训,可以节省学习成本。](https://github.com/pythonpeixun/article/blob/master/index.md)

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