https://blog.youkuaiyun.com/lemon_tree12138/article/details/51176153
这一篇对这个问题进行了比较详细的描述。下面是我写的高时间复杂度,低空间复杂度的python 脚本。性能一般,胜在原创。
class Solution(object):
def setZeroes(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: void Do not return anything, modify matrix in-place instead.
"""
NoneZero_X = []
NoneZero_Y = []
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if abs(matrix[i][j]) == 0:
break
if j == len(matrix[0]) - 1 and abs(matrix[i][j]) > 0 :
NoneZero_X.append(i)
for i in range(len(matrix[0])):
for j in range(len(matrix)):
print i,j
if abs(matrix[j][i]) == 0:
break
if j == len(matrix) - 1 and abs(matrix[j][i]) > 0 :
NoneZero_Y.append(i)
#print NoneZero_X
#print NoneZero_Y
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if i not in NoneZero_X or j not in NoneZero_Y:
matrix[i][j] = 0