描述
Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.
Example:
Input: s = "abcdefg", k = 2
Output: "bacdfeg"分析
这是一个字符串反转的问题，只是这里多了一个步长k的参数。如果前k个参数进行了反转，则后k个字符串不进行反转。因此我们可以设置一个标志位flag，如果为True，则对接下来k个字符串反转，否则保持原状。每k步对flag进行一次取反。
代码
class Solution:
def reverseStr(self, s, k):
"""
:type s: str
:type k: int
:rtype: str
"""
flag = False
temp = ""
for i in range(0, len(s), k):
flag = not flag
stop = i+k
if stop > len(s):
stop = len(s)
if flag:
temp += s[i:stop][::-1]
else:
temp += s[i:stop]
return temp优化：
看了下beats 100%的代码，以2k为步长，则每次迭代只需将反转之前的、反转的和反转之后的三部分加起来，即每2k个字符是一个子问题：
for idx in range(0, len(s), 2*k):
s = s[:idx] + s[idx:idx+k][::-1] + s[idx+k:]
return s