diff --git a/103. Scramble String/README.md b/103. Scramble String/README.md
new file mode 100644
index 0000000..f96a234
--- /dev/null
+++ b/103. Scramble String/README.md	
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+两个字符串 s1, s2, 问是否是 scramble. --> 定义问题的解为：`S(s1, s2)`, s1,s2 的长度为 n (若不等，直接 `return false;`).
+
+然后看给出的例子：
+
+	     gr|eat               rg|tae          ----- i == 2 (1)
+	     /    \               /    \
+	   g|r   e|at           r|g   ta|e        ----- i == 1 (2)
+	   / \   / \            / \    / \
+	  g   r e   a|t        r   g t|a  e       ----- i == 1 (3)
+	            / \              / \
+	           a   t            t   a
+
+可以来看看，这里面究竟有几种情况。按照上图所示的步骤进行分析：
+
+1. `S(s1, s2) = S(s1.substr(0,i), s2.substr(0,i)) && S(s1.substr(i), s2.substr(i))` (第一种可能)
+2. 重点看 `e|at` 与 `ta|e`, 可以推得: `S(s1, s2) = S(s1.substr(0,i), s2.substr(n-i)) && S(s1.substr(i), s2.substr(0, n-i))` (第二种可能)
+3. 可以并入 2. 但可以看出 S 最简单的解: `S("a", "a") == true` 以及 `S("t", "t") == true`. 即 `if (s1 == s2) return true;`
+
+综上三步，可以得到递归的公式：
+
+                 _________ true; 
+     S(s1, s2) _|s1 == s2
+                |      __ S(s1.substr(0,i), s2.substr(0,i)) && S(s1.substr(i), s2.substr(i))
+                |_ or |
+                      |__ S(s1.substr(0,i), s2.substr(n-i)) && S(s1.substr(i), s2.substr(n-i))
+
+
+显然 i 的范围是 [1 ,n-1]. 
+
+-----
+
+有了这个公式做支撑，我们基本看到了递归的雏形。也可以试着用这个来写出代码。
+
+```cpp
+
+if (s1.empty() || s2.empty() || s1.size() != s2.size()) return false;
+
+else if (s1 == s2) return true;
+
+
+for (size_t i=1; i<s1.size(); ++i)
+
+     if (isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i))) return true;
+
+     else if (isScramble(s1.substr(0,i), s2.substr(s2.size()-i)) && isScramble(s1.substr(i), s2.substr(0, s2.size()-i))) return true;
+
+return false;
+
+```
+
+六行代码基本已经从逻辑上解出了这道题。但如果去 LeetCode 上提交，会发现 TimeOut. 因为这样其实与枚举无异。如果遇到连字符都不尽相同的两个字符串，实在是浪费时间。
+
+于是我们考虑排除两种情况：
+
+1. s1 与 s2 中字符组成有差异，如 s1 中有 'g', s2 中却无。直接 `return false;`
+2. s1 与 s2 中字符出现的个数有差异，如 s1 中 'g' 出现两次，而 s2 中却只有一次。直接 `return false;`.
+
+根据这俩个条件，我们在第三行补充代码：
+
+```cpp
+for(auto c : s1)
+
+     if (s2.find_first_of(c) == string::npos) return false;
+
+     else if (std::count(s1.cbegin(), s1.cend(), c) != std::count(s2.cbegin(), s2.cend(), c)) return false;
+
+```
+
+然后提交，AC, 16ms. 问题顺利解决。
diff --git a/103. Scramble String/TEST.cc b/103. Scramble String/TEST.cc
new file mode 100644
index 0000000..b122d45
--- /dev/null
+++ b/103. Scramble String/TEST.cc	
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+#define CATCH_CONFIG_MAIN
+#include "../Catch/single_include/catch.hpp"
+#include "solution.h"
+
+TEST_CASE("Scramble String", "[isScramble]")
+{
+    Solution s;
+    REQUIRE(s.isScramble("great", "rgtae"));
+    REQUIRE_FALSE(s.isScramble("ccabcbabcbabbbbcbb", "bbbbabccccbbbabcba"));
+}
diff --git a/103. Scramble String/solution.h b/103. Scramble String/solution.h
new file mode 100644
index 0000000..100b392
--- /dev/null
+++ b/103. Scramble String/solution.h	
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+#include <string>
+using std::string;
+#include <algorithm>
+
+class Solution {
+public:
+    bool isScramble(string s1, string s2) {
+        if (s1.empty() || s2.empty() || s1.size() != s2.size()) return false;
+        else if (s1 == s2) return true;
+        for(auto c : s1)
+            if (s2.find_first_of(c) == string::npos) return false;
+            else if (std::count(s1.cbegin(), s1.cend(), c) != std::count(s2.cbegin(), s2.cend(), c)) return false;
+        for (size_t i=1; i<s1.size(); ++i)
+            if (isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i))) return true;
+            else if (isScramble(s1.substr(0,i), s2.substr(s2.size()-i)) && isScramble(s1.substr(i), s2.substr(0, s2.size()-i))) return true;
+        return false;
+    }
+};