作业（十三）——541. Reverse String II

题目

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"

Restrictions:

1. The string consists of lower English letters only.
2. Length of the given string and k will in the range [1, 10000]

思路

题目是将字符串每2k个字符反转前k个

我的做法是while循环每次减去k个字符，设置一个反转标志，判断这段是否反转。

代码

class Solution:
    def reverseStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        m=[]
        f=1
        while len(s)>k:
            if f==1:
                m.append(s[:k][::-1])
            else:
                m.append(s[:k])
            f=-f
            s=s[k:]
        if f==1:
            m.append(s[:][::-1])
        else:
            m.append(s[:])

        return ''.join(m)

结果