# 541. 反转字符串 II - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解
访问原文链接：[541. 反转字符串 II - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.to/zh/leetcode/541-reverse-string-ii)，体验更佳！
力扣链接：[541. 反转字符串 II](https://leetcode.cn/problems/reverse-string-ii), 难度等级：**简单**。
## LeetCode “541. 反转字符串 II”问题描述
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
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`。
```ruby
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。
```ruby
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)`.
## Python
```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
```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
```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#
```csharp
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
```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++
```cpp
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
```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)
}
```
## Other languages
```java
// Welcome to create a PR to complete the code of this language, thanks!
```
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原文链接：[541. 反转字符串 II - LeetCode Python/Java/C++/JS/C#/Go/Ruby 题解](https://leetcode.to/zh/leetcode/541-reverse-string-ii).
GitHub 仓库: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).