【LetMeFly】541.反转字符串 II:模拟
力扣题目链接:https://leetcode.cn/problems/reverse-string-ii/
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"
提示:
1 <= s.length <= 10
^4^s
仅由小写英文组成1 <= k <= 10
^4^
解题方法:模拟
使用l
从0
到len(s) - 1
遍历要翻转字符串的左端点,那么要翻转字符串的右端点就是min(l + r, len(s)) - 1
。每次l += 2*k
。
对于字符串s[l, r]
,如何翻转?若不能调用编程语言内置函数,可在l < r
时交换s[l]
和s[r]
并右移l
和左移r
各一次。
- 时间复杂度 O ( l e n ( s ) ) O(len(s)) O(len(s))
- 空间复杂度 O ( 1 ) O(1) O(1)
AC代码
C++
cpp
/*
* @Author: LetMeFly
* @Date: 2025-01-31 11:58:52
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-01-31 12:02:00
*/
class Solution {
private:
inline void reverse_(string& s, int l, int r) {
reverse(s.begin() + l, s.begin() + r);
}
public:
string reverseStr(string& s, int k) {
for (int l = 0; l < s.size(); l += k * 2) {
reverse_(s, l, min(l + k, (int)s.size()));
}
return s;
}
};
Python
python
'''
Author: LetMeFly
Date: 2025-01-31 12:02:40
LastEditors: LetMeFly.xyz
LastEditTime: 2025-01-31 12:08:02
'''
class Solution:
def reverseStr(self, s: str, k: int) -> str:
s = list(s)
for l in range(0, len(s), k * 2):
s[l:l + k] = s[l:l + k][::-1]
return ''.join(s)
Java
java
/*
* @Author: LetMeFly
* @Date: 2025-01-31 12:11:33
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-01-31 12:16:13
*/
class Solution {
private void reverse(char[] s, int l, int r) { // [l, r]
while (l < r) {
char temp = s[l];
s[l++] = s[r];
s[r--] = temp;
}
}
public String reverseStr(String s1, int k) {
char[] s = s1.toCharArray();
for (int l = 0; l < s.length; l += k * 2) {
reverse(s, l, Math.min(l + k, s.length) - 1);
}
// return s.toString(); // 不可!
return new String(s);
}
}
Go
go
/*
* @Author: LetMeFly
* @Date: 2025-01-31 12:17:35
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-01-31 12:18:54
*/
package main
import "slices"
func reverseStr(s1 string, k int) string {
s := []byte(s1)
for i := 0; i < len(s); i += k * 2 {
slices.Reverse(s[i:min(i + k, len(s))])
}
return string(s)
}
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Tisfy:https://letmefly.blog.csdn.net/article/details/145404901