## [面试题29. 顺时针打印矩阵](https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/) >Easy #### 思路 * 按着题目理解的方式进行遍历 * 一圈分成4个步骤，代码使用 `%4` 来区分步骤 * 一圈结束后，要向内圈缩进一格 * 遍历次数为m * n #### 代码 python3 ``` class Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: if len(matrix) == 0: return [] r = [] m = len(matrix) n = len(matrix[0]) count = 0 i = 0 j = 0 for k in range(m * n): #print((i,j)) r.append(matrix[i][j]) if count % 4 == 0: if j < (n - 1 - int(count / 4)): j +=1 else: i+=1 count += 1 elif count % 4 == 1: if i < (m - 1 - int(count / 4)): i +=1 else: j-=1 count +=1 elif count % 4 == 2: if j > int(count / 4): j-=1 else: i-=1 count +=1 else: if i> int(count/4) + 1: i-=1 else: j+=1 count +=1 return r ```