leetcode-python-矩阵置0

笨比方法，空间复杂度较高

1）逐行遍历，发现0记录列的位置，然后当前行全为0。最后再遍历一次，每行的对应列也置0

class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        row = list()
        column = list()
        for i in range(len(matrix)):
            flag = 0
            for j in range(len(matrix[0])):
                if matrix[i][j] == 0:
                    if j not in column:
                        column.append(j)
                    flag += 1
            if flag == 0:
                row.append(i)
            else:
                matrix[i] = [0]*len(matrix[0])
        for i in row:
            for j in column:
                matrix[i][j] = 0