Question

检查两个字符串是否是字谜。编写一个函数anagrams（s1，s2），给定两个字符串s1和s2，如果它们是，则返回True 字谜，否则为False，否则使用字典

如果len（s1）与len（s2）不同，则它们不是字谜
我找不到使用字典比较两个字符串的方法。

代码：

magick convert map.png -negate -morphology open disk:5 -negate result.png

Answer 1

您可以对每个字符串使用字典来计算每个不同字符的出现次数：

def anagrams(s1, s2):
    d = {}
    for s in s1, s2:
        d[s] = {}
        for c in s:
            d[s][c] = d[s].get(c, 0) + 1
    return d[s1] == d[s2]

Answer 2

如果要使用字典，请检查下面的代码

def anagrams(s1,s2):
    s = s1+s2 #for example s = asd + dsa = asddsa
    l = list(s) #[a,s,d,d,s,a]
    dic = dict(enumerate(l)) # {0: 'a', 1: 's', 2: 'd', 3: 'd', 4: 's', 5: 'a'}
    if len(dic)%2 == 1: #if the two strings are anagrams, the length of the combined strings should be even number
        return False
    else: # now we just compare the two ends of all keys, in the above example, we compare 0 and 5 / 1 and 4 / 2 and 3
        # Notice: the sum of i and the its corresponding party is the len of dic
        i = 0
        while i < len(dic)/2:
            if dic[i] != dic[len(dic)-1-i]:
                return False
                break

            else:
                i += 1
        return True

或者，您可以使用满足相同目的的双端队列功能。简单的逻辑是将拖车绳加在一起并比较两端

from collections import deque
def anagrams(s1,s2):
    s = s1+s2 # put them into one string and now we can simply compare if the far left and far right one is the same
    dq = deque(s) # apply deque function to it
    while len(dq) > 1: #if it equals to one
        if dq.popleft() != dq.pop():
            return False
    if len(dq) == 1:
        return False
    else:
        return True

Answer 3

您可以将单词加载到词典中并比较字典的排序值。

D1={}
D2={}

def anagrams(s1,s2):
    if len(s1)!=len(s2):
        return False

    else:
        elementNumber = 0
        for char in s1:                    #Load s1 into dictionary
            D1[elementNumber] = char
            elementNumber = elementNumber + 1

        elementNumber = 0
        for char in s2:                    #Load s2 into dictionary
            D2[elementNumber] = char
            elementNumber = elementNumber + 1

        print(sorted(D1.values()))         #Example output
        print(sorted(D2.values()))         #Example output

        if sorted(D1.values())==sorted(D2.values()): #Sort and compare
             return True
        else:
             return False

print("Anagrams: "+str(anagrams("Hello", "oHlel"))) #Returns True
print("Anagrams: "+str(anagrams("Hello", "xyzlo"))) #Returns False

Answer 4

如果您只是检查一个字谜，请尝试使用python的Counter对象。您只需要一行即可。

来自anagram with Counter

# Python code to check if two strings are 
# anagram 
from collections import Counter 

def anagram(input1, input2): 

    # Counter() returns a dictionary data 
    # structure which contains characters  
    # of input as key and their frequencies 
    # as it's corresponding value 
    return Counter(input1) == Counter(input2)

使用字典时比较2个字符串（字谜）

4 个答案: