leetcode剑指 Offer 29. 顺时针打印矩阵---同54题

剑指 Offer 29. 顺时针打印矩阵—同54题

class Solution(object):
    def spiralOrder(self, matrix):
        ans = []
        if not matrix:
            return ans
        #定义上下左右边界
        up, down, left, right = 0, len(matrix) - 1, 0, len(matrix[0]) - 1
        while True:
            for i in range(left, right + 1):   #[左-->右)
                ans.append(matrix[up][i])      #遵循：循环不变量原则，都是先闭后开
            up += 1
            if up > down:
                break
            for i in range(up, down + 1):      #[上-->下)
                ans.append(matrix[i][right])
            right -= 1
            if right < left:
                break
            for i in range(right, left - 1, -1):#[右-->左)
                ans.append(matrix[down][i])
            down -= 1
            if down < up:
                break
            for i in range(down, up - 1, -1):   #[下-->上)
                ans.append(matrix[i][left])
            left += 1
            if left > right:
                break
        return ans