题解 | #顺时针打印矩阵#

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param matrix int整型二维数组
# @return int整型一维数组
#
class Solution:
    def printMatrix(self, matrix):
        # write code here
        n = len(matrix)
        m = len(matrix[0])
        if not m or not n: # 特殊条件1，空
            return []
        if n == 1: # 特殊条件2，一行
            return matrix[0]
        if m == 1: # 特殊条件3，一列
            return [item[0] for item in matrix]
        flag = [[False] * m for _ in range(n)] # 其他情况
        direct = 0 # 表示当前正在游走的方向
        wherego = 0 # 表示接下来拐弯应去向何方
        i, j = 0, 0
        res = []
        while flag[i][j] == False:
            if wherego == 0: # 其实默认为0，向右
                while j < m - 1 and flag[i][j] == False:
                    res.append(matrix[i][j])
                    flag[i][j] = True # 走过的路定位False
                    j += 1 # 向右
                    direct = 1 # 当前游走位置为1
                    wherego = 1 # 接下来将去往1，这两个1含义不同，自己是自己的
            if wherego == 1:
                while i < n - 1 and flag[i][j] == False:
                    res.append(matrix[i][j])
                    flag[i][j] = True
                    i += 1
                    direct = 2
                    wherego = 2
            if wherego == 2:
                while j >= 1 and flag[i][j] == False:
                    res.append(matrix[i][j])
                    flag[i][j] = True
                    j -= 1
                    direct = 3
                    wherego = 3
            if wherego == 3:
                while i >= 1 and flag[i][j] == False:
                    res.append(matrix[i][j])
                    flag[i][j] = True
                    i -= 1
                    direct = 4
            if direct == 1: # 如果循环是从direct=1的方向跳出来的，那么接下来去向wherego=1的循环，以下同理
                j -= 1 # 对应循环跳出，i，j重新调整
                i += 1
                wherego = 1
            elif direct == 2:
                j -= 1
                i -= 1
                wherego = 2
            elif direct == 3:
                j += 1
                i -= 1
                wherego = 3
            else:
                i += 1
                j += 1
                wherego = 0
        return res