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算法:几个重要函数应用于一个题目
今天欣赏一个包含众多知识点的解法。题目是这样的:
给定两个字符串s1和s2,统计它们各自包含的各个小写字母的个数,然后互相比较。如果对某个字母,s1中出现的次数多,就打印“1:aaaa” (表示字母a在s1中出现了4次);如果s2中出现的次数多,就打印“2:aaaa”;如果次数相同,就打印“=:aaaa”。最后输出整个的比较结果,输出时先按次数排序,再按1-2-=排序,最后按字母排序。
举个例子:
s1 = "my&friend&Paul has heavy hats! &"
s2 = "my friend John has many many friends &"
mix(s1, s2) --> "2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss/1:l/1:t/1:u/1:v/2:o"
分析:
这个题目其实挺简单的。统计字母个数?遍历+字典轻松搞定。比较两个字典?逐个即可。最后输出的先按...排序,再按...排序,最后按...排序?有点麻烦,不过用itemgetter也可以做到。
本人参考两种解法,并比较两种解法的性能:
1 #-*- coding:utf-8 -*- 2 __author__ = ‘Administrator‘ 3 4 from collections import Counter 5 import re 6 import time 7 #s1 = "my&friend&Paul has heavy hats! &" 8 #s2 = "my friend John has many many friends &" 9 #str_mix(s1, s2) --> "2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss/1:l/1:t/1:u/1:v/2:o" 10 def str_min(s1,s2): 11 c1=Counter(s1) 12 c2=Counter(s2) 13 d=[] 14 key1=c1.keys(); 15 key2=c2.keys(); 16 17 for key in set(list(key1)+list(key2)): 18 if re.match("[a-z]",key): 19 if c1[key]>c2[key]: 20 d.append(("1",key,c1[key])) 21 elif c1[key]<c2[key]: 22 d.append(("2",key,c2[key])) 23 else: 24 d.append(("=",key,c1[key])) 25 26 rec=["{0}:{1}".format(v,s*n) for v,s,n in d] 27 return rec 28 29 def str_min2(s1,s2): 30 c1=Counter(filter(str.islower,s1)) 31 c2=Counter(filter(str.islower,s2)) 32 d=[] 33 for key in set(list(c1.keys())+list(c2.keys())): 34 n1=c1.get(key,0) 35 n2=c2.get(key,0) 36 if n1>0 or n2>0: 37 if n1>n2: 38 d.append(("1",key,n1)) 39 elif n1<n2: 40 d.append(("2",key,n2)) 41 else: 42 d.append(("=",key,n1)) 43 rec=["{}:{}".format(v,s*n) for v,s,n in d] 44 return rec 45 46 if __name__=="__main__": 47 s1 = "my&friend&Paul has heavy hats! &" 48 s2 = "my friend John has many many friends &" 49 t0=time.time() 50 for i in range(100000): 51 rec=str_min(s1,s2) 52 print(‘/‘.join(sorted(rec,key=lambda s:(-len(s),s)))) 53 t1=time.time()-t0 54 print(t1) 55 print("-"*100) 56 t0=time.time() 57 for i in range(100000): 58 rec=str_min2(s1,s2) 59 print(‘/‘.join(sorted(rec,key=lambda s:(-len(s),s)))) 60 t1=time.time()-t0 61 print(t1) 62 63
输出:
2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss/1:l/1:t/1:u/1:v/2:o
14.07480502128601
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2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss/1:l/1:t/1:u/1:v/2:o
8.899508953094482
算法:几个重要函数应用于一个题目