力扣链接：541. 反转字符串 II，难度等级：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 <= 10000sconsists of only lowercase English letters.1 <= k <= 10000
思路
- 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。 - 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。 - 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
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)
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)
}