剑指 Offer 29. 顺时针打印矩阵

剑指 Offer 29. 顺时针打印矩阵

描述

略

思路

画好图，定义好(x1, y1, x2, y2)的物理意义

• 注意循环条件
• 注意终止位置
• 更新内外层
  
  图片来自https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/solution/shun-shi-zhen-da-yin-ju-zhen-by-leetcode-solution/

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        if not matrix:
            return []
        ret = []
        x1, y1, x2, y2 = 0, 0, len(matrix) - 1, len(matrix[0]) - 1
        while x1 <= x2 and y1 <= y2:
            for i in range(y1, y2 + 1):
                ret.append(matrix[x1][i])
            for i in range(x1 + 1, x2 + 1):
                ret.append(matrix[i][y2])
            if x1 < x2 and y1 < y2:
                for i in range(y2 - 1, y1, -1):
                    ret.append(matrix[x2][i])
                for i in range(x2, x1, -1):
                    ret.append(matrix[i][y1])
            x1, y1, x2, y2 = x1 + 1, y1 + 1, x2 - 1, y2 - 1
        return ret