Reverse string II.

Author: recherst

Easy/541.Reverse String II.playground/Contents.swift

@@ -0,0 +1,52 @@
+/**
+ Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.
+
+ If there are fewer 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 leave the other as original.
+
+  
+
+ Example 1:
+ Input: s = "abcdefg", k = 2
+ Output: "bacdfeg"
+ 
+ Example 2:
+ Input: s = "abcd", k = 2
+ Output: "bacd"
+  
+
+ Constraints:
+ - 1 <= s.length <= 10^4
+ - s consists of only lowercase English letters.
+ - 1 <= k <= 10^4
+ */
+class Solution {
+    func reverseStr(_ s: String, _ k: Int) -> String {
+        if s.count <= k {
+            return String(s.reversed())
+        } else {
+            var result = s
+            let quotient = s.count / (2 * k)
+            let remiander = s.count % (2 * k)
+            if quotient > 0 {
+                for i in 0..<quotient {
+                    let subString = String(s[s.index(s.startIndex, offsetBy: i * 2 * k)..<s.index(s.startIndex, offsetBy: i * 2 * k + k)].reversed())
+                    result.replaceSubrange(s.index(s.startIndex, offsetBy: i * 2 * k)..<s.index(s.startIndex, offsetBy: i * 2 * k + k), with: subString)
+                }
+            }
+            if remiander > 0 {
+                if remiander >= k {
+                    let subString = String(s[s.index(s.startIndex, offsetBy: quotient * 2 * k)..<s.index(s.startIndex, offsetBy: quotient * 2 * k + k)].reversed())
+                    result.replaceSubrange(s.index(s.startIndex, offsetBy: quotient * 2 * k)..<s.index(s.startIndex, offsetBy: quotient * 2 * k + k), with: subString)
+                } else {
+                    let subString = String(s[s.index(s.startIndex, offsetBy: quotient * 2 * k)...s.index(s.startIndex, offsetBy: s.count - 1)].reversed())
+                    result.replaceSubrange(s.index(s.startIndex, offsetBy: quotient * 2 * k)...s.index(s.startIndex, offsetBy: s.count - 1), with: subString)
+                }
+            }
+            return result
+        }
+    }
+}
+
+let s = Solution()
+let r = s.reverseStr("abcd", 2)
+print(r)

Easy/541.Reverse String II.playground/contents.xcplayground

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+<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
+<playground version='5.0' target-platform='ios' buildActiveScheme='true' importAppTypes='true'>
+    <timeline fileName='timeline.xctimeline'/>
+</playground>

Easy/541.Reverse String II.playground/playground.xcworkspace/contents.xcworkspacedata

@@ -107,6 +107,7 @@
 103. [Fibonacci Number](https://github.com/recherst/leetcode-algtorithm/blob/main/Easy/509.Fibonacci%20Number.playground/Contents.swift)
 104. [Detect Capital](https://github.com/recherst/leetcode-algtorithm/blob/main/Easy/520.Detect%20Captial.playground/Contents.swift)
 105. [Longest Uncommon Subsequence I](https://github.com/recherst/leetcode-algtorithm/blob/main/Easy/521.Longest%20Uncommon%20Subsequence%20I.playground/Contents.swift)
+106. [Reverse String II](https://github.com/recherst/leetcode-algtorithm/blob/main/Easy/541.Reverse%20String%20II.playground/Contents.swift)
 
 #### Medium