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diff --git a/assets/problems/54.spiral-matrix.jpg b/assets/problems/54.spiral-matrix.jpg
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diff --git a/daily/2019-07-29.md b/daily/2019-07-29.md
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+# 毎日一题 - 54.Spiral Matrix
+
+## 信息卡片
+
+- 时间：2019-07-29
+- 题目链接：https://leetcode.com/problems/spiral-matrix/
+- tag：`Array` `Matrix`
+
+## 题目描述
+
+```
+Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
+
+Example 1:
+
+  Input:
+    [
+      [ 1, 2, 3 ],
+      [ 4, 5, 6 ],
+      [ 7, 8, 9 ]
+    ]
+  Output: [1,2,3,6,9,8,7,4,5]
+Example 2:
+
+  Input:
+    [
+      [1, 2, 3, 4],
+      [5, 6, 7, 8],
+      [9,10,11,12]
+    ]
+  Output: [1,2,3,4,8,12,11,10,9,5,6,7]
+```
+
+## 参考答案
+
+1. 剥洋葱，row->col->row->col 为一次；
+2. row->col、col->row 的切换都伴随读取的初始位置的变化；
+3. 结束条件是row头>row尾或者col顶>col底
+
+![剥洋葱](../assets/problems/54.spiral-matrix.jpg)
+
+时间复杂度O(m*n), 空间复杂度O(1)
+
+参考JavaScript代码：
+
+```js
+/**
+ * @param {number[][]} matrix
+ * @return {number[]}
+ */
+var spiralOrder = function(matrix) {
+    if(matrix.length === 0) return [];
+    let rowT = 0; // 行顶
+    let rowB = matrix.length - 1; // 行底
+    let colL = 0; // 列左
+    let colR = matrix[0].length - 1; // 列右
+    let result =  [];
+    // 顺序是行、列、行、列；每次切换，读取的初始位置都会变化1(+/- 1)
+    while (colL <= colR && rowT <= rowB) {
+      for (let a = colL; a <= colR; a++) {
+        result.push(matrix[rowT][a]);
+      }
+      rowT++;
+      for (let b = rowT; b <= rowB; b++) {
+        result.push(matrix[b][colR]);
+      }
+      colR--;
+      for (let c = colR; c >= colL && rowB >= rowT; c--) {
+        result.push(matrix[rowB][c]);
+      }
+      rowB--;
+      for (let d = rowB; d >= rowT && colR >= colL; d--) {
+        result.push(matrix[d][colL]);
+      }
+      colL++;
+    }
+    return result;
+};
+```
+
+代码只有一个for循环的方式，操作方向
+例如
+>  1  2  3  4  5
+>  6  7  8  9 10
+> 11 12 13 14 15
+>
+> 对上面矩阵遍历时的操作
+>
+> 向右5次(算上从左侧第一次进入)
+> 向下2次
+> 向左4次
+> 向上1次
+> 向右3次
+> 向下0次 -- 结束
+
+方向有四个，right、down、left、up
+四个方向又分两类，水平(right,left)和垂直(down，up)
+而在两类方向上的移动最值是 水平n, 垂直m;
+在遍历过程中，根据`方向切换`来减小n/m从而缩小两类方向的移动最值直到结束
+四个方向可以用二维数组来表示[ [0, 1], [1, 0], [0, -1], [-1, 0] ]
+两类方向各自的初始最大值是[n, m-1]
+当 n == 0 || m == 0 表示元素已经全部遍历完
+
+这种写法省去了代码中的for循环，但是while循环次数却增多了；复杂度没有变化
+时间复杂度O(m*n), 空间复杂度O(1)
+
+参考JavaScript代码：
+
+```js
+/**
+ * @param {number[][]} matrix
+ * @return {number[]}
+ * 一个for循环,但while变多了
+ */
+var spiralOrder = function(matrix) {
+    if(matrix.length === 0) return [];
+    let m = matrix.length;
+    let n = matrix[0].length;
+    let result = [];
+    const dirs = [[0, 1], [1, 0], [0, -1], [-1, 0]] // 控制方向的数组
+    // 元素坐标row,col;
+    let row = 0;
+    let col = -1;
+    let steps = [n, m-1]
+    let dir = 0; // 初始方向
+    while(steps[dir%2]) {
+      for(let i = 0; i < steps[dir%2]; i++) {
+        // 方向的改变的效果，row/col能增能减
+        row += dirs[dir][0]; col += dirs[dir][1];
+        result.push(matrix[row][col])
+      }
+      steps[dir%2]--; // 移动极值缩小
+      dir = (dir+1)%4; // 方向改变
+    }
+    return result;
+};
+```
+
+## 优秀解答
+
+> 暂缺
+
+## 参考
+- @stellari [A concise C++ implementation based on Directions](https://leetcode.com/problems/spiral-matrix/discuss/20573/A-concise-C%2B%2B-implementation-based-on-Directions)
diff --git a/daily/README.md b/daily/README.md
index 3dad09314..d9d4db7e9 100644
--- a/daily/README.md
+++ b/daily/README.md
@@ -173,3 +173,9 @@ tag：`Math`
 tag：`数据压缩`
 
 时间: 2019-07-26
+
+### [54.Spiral Matrix](./2019-07-29.md)
+
+tag：`Array` `Matrix`
+
+时间: 2019-07-29