向前和向后遍历字符串，提取交替字符

Question

I'm studying python and there's a lab I can't seem to crack. We have a line eg shacnidw , that has to be transformed to sandwich . I somehow need to iterate with a for loop and pick the letters with odd indexes first, followed by backward even indexes. Like pick a letter with index 1,3,5,7,8,6,4,2.

It looks pretty obvious to use a list or slices, but we aren't allowed to use these functions yet. I guess the question is just how do I do it?

Answer 1

I hope I am understanding your question right. Check the below code.

s = 'shacnidw'
s_odd = ""
s_even = ""
for i in range(len(s)):
  if i%2 == 0:
    s_even += s[i]
for i in range(len(s), 0, -1):
  if i%2 == 1:
    s_odd += s[i]

print(s_even + s_odd)

I hope it might help.

Answer 2

Programming is all about decomposing complex problems into simpler ones. Try breaking it down into smaller steps.

1. First, can you generate the numbers 1,3,5,7 in a for loop?
2. Next, can you generate 8,6,4,2 in a second loop?

Tackling those two steps ought to get you on the right track.

Answer 3

Your text looks to be in a specific pattern that can be achieved using the following syntax.

1, 3, 5, 7: for n in range(1, size, 2): steps of 2 in forward direction

8, 6, 4, 2: for m in range(size, 0, -2): steps of 2 in reverse direction

By building the string with the above indices we can easily arrive at the intended result.

string = 'shacnidw'
result = ''
size = len(string) - 1

for n in range(0, size, 2):
    result += string[n]

for m in range(size, 0, -2):
    result += string[m]

print(result)

Result

sandwich