翻转字符串

LRH1993 · LRH1993 · commit 9d0d6f87ca79 · 2017-10-13T09:56:27.000+08:00
diff --git a/SUMMARY.md b/SUMMARY.md
@@ -200,6 +200,7 @@
     - [Two Sum](/algorithm/LeetCode/Array/Two-Sum.md)
   - [String](/algorithm/LeetCode/String.md)
     - [Restore IP Addresses](/algorithm/LeetCode/String/ip.md)
+    - [Rotate String](/algorithm/LeetCode/String/Rotate-String.md)
 
 ## 设计模式
 
diff --git a/algorithm/LeetCode/String/Rotate-String.md b/algorithm/LeetCode/String/Rotate-String.md
@@ -0,0 +1,58 @@
+## 一、题目
+
+> Given a string and an offset, rotate string by offset. (rotate from left to right)
+>
+> Example
+>
+> Given `"abcdefg"`.
+>
+> offset=0 => "abcdefg"
+> offset=1 => "gabcdef"
+> offset=2 => "fgabcde"
+> offset=3 => "efgabcd"
+>
+> Challenge
+>
+> Rotate in-place with O(1) extra memory.
+
+给定一个字符串和一个偏移量，根据偏移量旋转字符串(从左向右旋转)
+
+## 二、解题思路
+
+常见的翻转法应用题，仔细观察规律可知翻转的分割点在从数组末尾数起的offset位置。先翻转前半部分，随后翻转后半部分，最后整体翻转。
+
+## 三、解题代码
+
+```java
+public class Solution {
+    /*
+     * param A: A string
+     * param offset: Rotate string with offset.
+     * return: Rotated string.
+     */
+    public char[] rotateString(char[] A, int offset) {
+        if (A == null || A.length == 0) {
+            return A;
+        }
+
+        int len = A.length;
+        offset %= len;
+        reverse(A, 0, len - offset - 1);
+        reverse(A, len - offset, len - 1);
+        reverse(A, 0, len - 1);
+
+        return A;
+    }
+
+    private void reverse(char[] str, int start, int end) {
+        while (start < end) {
+            char temp = str[start];
+            str[start] = str[end];
+            str[end] = temp;
+            start++;
+            end--;
+        }
+    }
+}
+```
+
