题解 | #螺旋矩阵#

模拟遍历过程，注意边界条件

#
# 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 spiralOrder(self , matrix: List[List[int]]) -> List[int]:
        # write code here
        if not matrix: return []
        i = j = 0
        ans = []
        while 1:
            if matrix[i][j] == 101:
                break
            while i >= 0 and j >= 0 and i < len(matrix) and j < len(matrix[0]) and matrix[i][j] != 101:
                ans.append(matrix[i][j])
                matrix[i][j] = 101
                if j + 1 < len(matrix[0]) and matrix[i][j+1] != 101:
                    j = j+1
                else: break
            i = i+1
            while i >= 0 and j >= 0 and i < len(matrix) and j < len(matrix[0]) and matrix[i][j] != 101:
                ans.append(matrix[i][j])
                matrix[i][j] = 101
                if i + 1 < len(matrix) and matrix[i+1][j] != 101:
                    i = i+1
                else: break
            j = j-1
            while i >= 0 and j >= 0 and i < len(matrix) and j < len(matrix[0]) and matrix[i][j] != 101:
                ans.append(matrix[i][j])
                matrix[i][j] = 101
                if j - 1 >=0 and matrix[i][j-1] != 101:
                    j = j-1
                else: break
            i = i-1
            while i >= 0 and j >= 0 and i < len(matrix) and j < len(matrix[0]) and matrix[i][j] != 101:
                ans.append(matrix[i][j])
                matrix[i][j] = 101
                if i - 1 >=0 and matrix[i-1][j] != 101:
                    i = i-1
                else: break
            j = j+1
        return ans