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    力扣链接：541. 反转字符串 II，难度：简单。

    给定一个字符串 s 和一个整数 k，从字符串开头算起，每计数至 2k 个字符，就反转这 2k 字符中的前 k 个字符。

    • 如果剩余字符少于 k 个，则将剩余字符全部反转。
    • 如果剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符，其余字符保持原样。

    示例 1:

    输入: s = "abcdefg", k = 2

    输出: bacdfeg

    示例 2:

    输入: s = "abcd", k = 2

    输出: bacd

    约束:

    • 1 <= s.length <= 10000
    • s consists of only lowercase English letters.
    • 1 <= k <= 10000

    思路
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    1. 题目没有要求原地反转，所以用一个新的字符串result作为返回值，操作起来容易些。
    2. 在循环中，步进值用k比2k更方便，因为如果用2k，还是要再用k做判断。
    3. 要求只反转2k字符中前k个字符，所以需要一个布尔类型变量should_reverse作为是否要反转判断条件。

    步骤

    1. 用一个新的字符串result作为返回值。在循环中，步进值用k。

      result = ''
index = 0

while index < s.size
  k_chars = s[index...index + k]
  result += k_chars
  index += k
end

return result

    2. 用布尔类型变量should_reverse作为是否要反转判断条件，并只反转2k字符中前k个字符。

      result = ''
should_reverse = true # 1
index = 0

while index < s.size
  k_chars = s[index...index + k]

  if should_reverse # 2
    result += k_chars.reverse # 3
  else # 4
    result += k_chars
  end

  index += k
  should_reverse = !should_reverse # 5
end

return result

    复杂度

    时间复杂度

    O(N)

    空间复杂度

    O(N)

    题解语言： Python Java JavaScript C# Ruby C++ Go

    Python # 

    class Solution:
    def reverseStr(self, s: str, k: int) -> str:
        result = ''
        should_reverse = True
        index = 0

        while index < len(s):
            k_chars = s[index:index + k]

            if should_reverse:
                result += k_chars[::-1]
            else:
                result += k_chars

            index += k
            should_reverse = not should_reverse

        return result

    Java # 

    class Solution {
    public String reverseStr(String s, int k) {
        var result = new StringBuffer();
        var shouldReverse = true;
        var index = 0;

        while (index < s.length()) {
            var kChars = s.substring(index, Math.min(index + k, s.length()));

            if (shouldReverse) {
                result.append(new StringBuffer(kChars).reverse());
            } else {
                result.append(kChars);
            }

            index += k;
            shouldReverse = !shouldReverse;
        }

        return result.toString();
    }
}

    JavaScript # 

    var reverseStr = function (s, k) {
  let result = ''
  let shouldReverse = true
  let index = 0

  while (index < s.length) {
    const kChars = s.substr(index, k)

    if (shouldReverse) {
      result += [...kChars].reverse().join('')
    } else {
      result += kChars
    }

    index += k
    shouldReverse = !shouldReverse
  }

  return result
};

    C# # 

    public class Solution
{
    public string ReverseStr(string s, int k)
    {
        string result = "";
        bool shouldReverse = true;
        int index = 0;

        while (index < s.Length)
        {
            string kChars = s[index..Math.Min(index + k, s.Length)];

            if (shouldReverse)
            {
                result += new string(kChars.Reverse().ToArray());
            }
            else
            {
                result += kChars;
            }

            index += k;
            shouldReverse = !shouldReverse;
        }

        return result;
    }
}

    Ruby # 

    def reverse_str(s, k)
  result = ''
  should_reverse = true
  index = 0

  while index < s.size
    k_chars = s[index...index + k]

    if should_reverse
      result += k_chars.reverse
    else
      result += k_chars
    end

    index += k
    should_reverse = !should_reverse
  end

  result
end

    C++ # 

    class Solution {
public:
    string reverseStr(string s, int k) {
        string result = "";
        bool shouldReverse = true;
        int index = 0;

        while (index < s.length()) {
            auto kChars = s.substr(index, k);

            if (shouldReverse) {
                reverse(kChars.begin(), kChars.end());
            }

            result += kChars;
            index += k;
            shouldReverse = !shouldReverse;
        }

        return result;
    }
};

    Go # 

    func reverseStr(s string, k int) string {
    var result []rune
    shouldReverse := true
    index := 0

    for index < len(s) {
        end := index + k
        if end > len(s) {
            end = len(s)
        }
        kChars := []rune(s[index:end])

        if shouldReverse {
            for i, j := 0, len(kChars) - 1; i < j; i, j = i + 1, j - 1 {
                kChars[i], kChars[j] = kChars[j], kChars[i]
            }
        }

        result = append(result, kChars...)
        index += k
        shouldReverse = !shouldReverse
    }

    return string(result)
}
    题解语言： Python Java JavaScript C# Ruby C++ Go

    其它语言

    欢迎贡献代码到我们的 GitHub 仓库，非常感谢！本题解位置在 541. 反转字符串 II。
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