Create 54. 螺旋矩阵.md

Author: CompetitiveLin

Simulation/54. 螺旋矩阵.md

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+#### 54. 螺旋矩阵
+
+难度：中等
+
+---
+
+给你一个 `m` 行 `n` 列的矩阵 `matrix` ，请按照  **顺时针螺旋顺序**  ，返回矩阵中的所有元素。
+
+ **示例 1：** 
+
+![](https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg)
+```
+输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
+输出：[1,2,3,6,9,8,7,4,5]
+```
+
+ **示例 2：** 
+
+![](https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg)
+```
+输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
+输出：[1,2,3,4,8,12,11,10,9,5,6,7]
+```
+
+ **提示：** 
+
+*   `m == matrix.length`
+*   `n == matrix[i].length`
+*   `1 <= m, n <= 10`
+*   `-100 <= matrix[i][j] <= 100`
+
+---
+
+模拟：
+
+用若干个变量分别表示下一步走的方向，上下左右四个方向的边界以及当前点的坐标。再分别考虑当前点到达角落时后续的移动情况。
+
+```Go
+func spiralOrder(matrix [][]int) []int {
+    left, right, up, down := 0, len(matrix[0]) - 1, 0, len(matrix) - 1
+    first, second := 0, 0
+    flag, sum := 1, (right + 1) * (down + 1)
+    res := make([]int, sum)
+    for cnt := 0; cnt < sum; cnt++ {
+        res[cnt] = matrix[first][second]
+        if flag == 1 {
+            if second == right{
+                flag = 2
+                up++
+                first++
+            }else {
+                second++
+            }
+        } else if flag == 2 {
+            if first == down {
+                flag = 3
+                right--
+                second--
+            } else {
+                first++
+            }
+        } else if flag == 3 {
+            if second == left {
+                flag = 4
+                down--
+                first--
+            } else {
+                second--
+            }
+        } else if flag == 4 {
+            if first == up {
+                flag = 1
+                left++
+                second++
+            } else {
+                first--
+            }
+        }
+    }
+    return res
+}
+```