0541. Reverse String I I

# 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]

## 题目大意 #

• 该字符串只包含小写的英文字母。
• 给定字符串的长度和 k 在[1, 10000]范围内。

## 解题思路 #

• 要求按照一定规则反转字符串：每 `2 * K` 长度的字符串，反转前 `K` 个字符，后 `K` 个字符串保持不变；对于末尾不够 `2 * K` 的字符串，如果长度大于 `K`，那么反转前 `K` 个字符串，剩下的保持不变。如果长度小于 `K`，则把小于 `K` 的这部分字符串全部反转。
• 这一题是简单题，按照题意反转字符串即可。

## 代码 #

``````
package leetcode

func reverseStr(s string, k int) string {
if k > len(s) {
k = len(s)
}
for i := 0; i < len(s); i = i + 2*k {
if len(s)-i >= k {
ss := revers(s[i : i+k])
s = s[:i] + ss + s[i+k:]
} else {
ss := revers(s[i:])
s = s[:i] + ss
}
}
return s
}

func revers(s string) string {
bytes := []byte(s)
i, j := 0, len(bytes)-1
for i < j {
bytes[i], bytes[j] = bytes[j], bytes[i]
i++
j--
}
return string(bytes)
}

``````

Apr 8, 2023