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website/content/ChapterFour/0541.Reverse-String-II.md

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+# [541. Reverse String II](https://leetcode.com/problems/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
+
+给定一个字符串和一个整数 k，你需要对从字符串开头算起的每个 2k 个字符的前k个字符进行反转。如果剩余少于 k 个字符，则将剩余的所有全部反转。如果有小于 2k 但大于或等于 k 个字符，则反转前 k 个字符，并将剩余的字符保持原样。
+
+要求:
+
+- 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
+
+```go
+
+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
+}
+
+```