diff --git a/119. Spiral Matrix/README.md b/119. Spiral Matrix/README.md
new file mode 100644
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+++ b/119. Spiral Matrix/README.md	
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+m * n 的矩阵按螺旋顺序转为数组：
+
+    1 2 3
+    4 5 6     -->     1 2 3 6 9 8 7 4 5
+    7 8 9
+
+我想到的是，直接用下标作为限制。m 行，n 列，即范围:
+
+- row: 0 ~ m
+- col: 0 ~ n
+
+首先是列递增，(0,0), (0,1), (0,2). 然后行递增，(1,2), (2,2). 再后是列递减，(2,1), (2,0). 和行递减，(1,0).
+最后进入下一螺旋：(1,1). 到此为止，全部数组都迭代出来。
+
+可以发现的规律是，每一次螺旋，都按照以下顺序：
+
+1. 列递增 (row 降到最低)
+2. 行递增 (col 增到最高)
+3. 列递减 (row 增到最高)
+4. 行递减 (col 降到最低)
+
+写成 `for` 循环的形式，则为：
+
+```cpp
+for (; nMin<mMax && nMin<nMax; --mMax, --nMax, ++mMin, ++nMin) {
+    for (int i=nMin; i<nMax; ++i)
+        ret.push_back(matrix[mMin][i]);
+    for (int i=mMin+1; i<mMax; ++i)
+        ret.push_back(matrix[i][nMax-1]);
+    for (int i=nMax-2; i>=nMin; --i)
+        ret.push_back(matrix[mMax-1][i]);
+    for (int i=mMax-2; i>mMin; --i)
+        ret.push_back(matrix[i][nMin]);
+}
+```
+
+除此之外，需要增加一些边界条件判断：
+
+- `n = matrix[0].size();` // 要提前判断 matrix.empty()
+- 第三个 `for` 循环要避免重复，如仅有一行的情况，列递增之后，没有行递增，直接就列递减了，那必然重复。判断 `mMax-1 != mMin`.
+- 第四个 `for` 循环同理，避免行重复，判断 `nMax-1 != nMin`.
diff --git a/119. Spiral Matrix/TEST.cc b/119. Spiral Matrix/TEST.cc
new file mode 100644
index 0000000..4b753c6
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+++ b/119. Spiral Matrix/TEST.cc	
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+#define CATCH_CONFIG_MAIN
+#include "../Catch/single_include/catch.hpp"
+#include "solution.h"
+
+TEST_CASE("Spiral Matrix", "spiralOrder")
+{
+    Solution s;
+    SECTION( "Example" )
+    {
+        std::vector<std::vector<int>> matrix = {
+            {1,2,3},
+            {4,5,6},
+            {7,8,9}
+
+        };
+        std::vector<int> ret{1,2,3,6,9,8,7,4,5};
+        REQUIRE(s.spiralOrder(matrix) == ret);
+    }
+    SECTION( "One line" )
+    {
+        std::vector<std::vector<int>> matrix = {{6,9,7}};
+        std::vector<int> ret{6,9,7};
+        REQUIRE(s.spiralOrder(matrix) == ret);
+    }
+    SECTION( "Two line" )
+    {
+        std::vector<std::vector<int>> matrix = {
+            {1,2,3,4,5,6,7,8,9,10},
+            {11,12,13,14,15,16,17,18,19,20}
+        };
+        std::vector<int> ret{1,2,3,4,5,6,7,8,9,10,20,19,18,17,16,15,14,13,12,11};
+        REQUIRE(s.spiralOrder(matrix) == ret);
+    }
+    SECTION( "One Column" )
+    {
+        std::vector<std::vector<int>> matrix = {{7}, {9}, {6}};
+        std::vector<int> ret{7,9,6};
+        REQUIRE(s.spiralOrder(matrix) == ret);
+    }
+}
diff --git a/119. Spiral Matrix/solution.h b/119. Spiral Matrix/solution.h
new file mode 100644
index 0000000..441137e
--- /dev/null
+++ b/119. Spiral Matrix/solution.h	
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+#include <vector>
+using std::vector;
+
+class Solution {
+public:
+    vector<int> spiralOrder(vector<vector<int> > &matrix) {
+        vector<int> ret;
+        if (matrix.empty()) return ret;
+        int mMax = matrix.size();
+        int nMax = matrix[0].size();
+        int mMin = 0, nMin = 0;
+        ret.reserve(mMax*nMax);
+        for (; nMin<mMax && nMin<nMax; --mMax, --nMax, ++mMin, ++nMin) {
+            for (int i=nMin; i<nMax; ++i)
+                ret.push_back(matrix[mMin][i]);
+            for (int i=mMin+1; i<mMax; ++i)
+                ret.push_back(matrix[i][nMax-1]);
+            if (mMax-1 == mMin) break;
+            for (int i=nMax-2; i>=nMin; --i)
+                ret.push_back(matrix[mMax-1][i]);
+            if (nMax-1 == nMin) break;
+            for (int i=mMax-2; i>mMin; --i)
+                ret.push_back(matrix[i][nMin]);
+        }
+        return ret;
+    }
+};