如何仅使用 While 循环和条件按字母顺序排列字符串？

Question

Given the String "aaababababab" (Only 'a' or 'b' allowed): How do I order the string alphabetically to be like "aaaaaaabbbbb" by only using a while loop?

This is my current attempt:

word = "ababababaabab"
wordfixed = ""
i = 0

while i < len(word):
    if word[i] == "a":
        wordfixed += "a"
        i += 1

    elif word[i] == "b":
        wordfixed += (word[i])
        i += 1

print(wordfixed)

But it simply prints the original word back. How can I get it to print the desired output?

Answer 1

Using your current idea, you could just append to one end or the other depending on the value of the letter. This would allow a to accumulate at the start and b to build off the end.

word = "ababababaabab"
wordfixed = ""
i = 0

while i < len(word):
    if word[i] == "a":
        wordfixed = word[i] + wordfixed 
    elif word[i] == "b":
        wordfixed = wordfixed + word[i]
    i += 1


print(wordfixed)
# aaaaaaabbbbbb

Answer 2

You can create two "helper" variables: one to accumulate the a s, and one for the b s. Then in the end just concatenate them in the right order:

word = "ababababaabab"
wordfixed = ""
wordfixed2 = ""
wordfixed3 = ""
i = 0

while i < len(word):
    if word[i] == "a":
        wordfixed2 += "a"
    elif word[i] == "b":
        wordfixed3 += "b"
    i += 1

wordfixed = wordfixed2 + wordfixed3
print(wordfixed)

I think that's the right way to do it.

Answer 3

Your code does not work because you add the letter that you currently have at the end - so it will be the same afterwards.

You can instead count amount of 'a' 's while looping through the string once (for loop would be better suited but while can be used as well) - then print the remaining 'b' :

word = "ababababaabab"
lw = len(word)
i = 0 
a_s= 0  
while i < lw:
    if word[i] == 'a':
        print("a", end="")  # dont end the line quite yet
        a_s += 1 
    i += 1

# print b's for remaining wordlength minus printed a's
print('b' * (lw-a_s))

or better with for loop:

a_s = 0
for letter in word:
    if letter == "a":
        a_s +=1
        print(letter, end="")
print('b' * (lw-a_s))

Answer 4

I'll go with @Mark_Meyer solution but if you don't really need to use only the while loop, here is an easy way to go :

word = "ababababaabab"
print(word.count('a')*'a' + word.count('b')*'b')
# aaaaaaabbbbbb