华为 笔试
前两题比较简单，就不说了。
第三题是求所有依赖循环的，一开始是想做DFS找有向图中的所有环的，但是浪费了很长时间没写出来，后来就打算用生成式做一个递归。
只写了个思路，后面没debug，所以得了0分。
然后结束后自己debug了一下，也不知道能过多少测试样例了，就提供下自己的思路吧
#华为笔试##笔试题目##华为#class Solution:
def findpath(self, relys_dict, start_file, stop_file):
if start_file in relys_dict:
for file in relys_dict[start_file]:
if file != stop_file:
for f in self.findpath(relys_dict, file, stop_file):
res = [start_file] + f
yield res
else:
yield [start_file, stop_file]
if __name__ == '__main__':
res = ['a.h:b.h c.h d.h','b.h:c.h d.h','c.h:d.h','d.h:a.h', 'headfile is :a.h']
headfile = res[-1].split(':')[-1]
relyfile = res[:-1]
relyfile = [x.split(':') for x in relyfile]
relydict = dict(relyfile)
for k in relydict:
relydict[k] = relydict[k].split(' ')
a = Solution()
flag = False
for x in a.findpath(relydict, headfile, headfile):
if x[-1] == headfile:
flag = True
break
if not flag:
print('none loop include', headfile)
else:
print('Bad coding -- loop include as bellow:')
for x in a.findpath(relydict, headfile, headfile):
if x[-1]==headfile:
for i in range(len(x)-1):
if i != len(x)-2:
print(x[i], end = ' ')
else:
print(x[i])
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