Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
Example 1:
Input:
[
[1,1,1],
[1,0,1],
[1,1,1]
]
Output:
[
[1,0,1],
[0,0,0],
[1,0,1]
]
Example 2:
Input:
[
[0,1,2,0],
[3,4,5,2],
[1,3,1,5]
]
Output:
[
[0,0,0,0],
[0,4,5,0],
[0,3,1,0]
]
Follow up:
A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?
class Solution(object):
def setZeroes(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: None Do not return anything, modify matrix in-place instead.
"""
m = len(matrix[0])
n = len(matrix)
if m == 0 or n == 0:return
row_col = []
for i in range(n):
for j in range(m):
if matrix[i][j] == 0:
row_col.append([i, j])
length = len(row_col)
for i in range(length):
r = row_col[i][0]
c = row_col[i][1]
for k in range(m):
matrix[r][k] = 0
for l in range(n):
matrix[l][c] = 0
return matrix
'''
if len(matrix)!=0:
tag=0
tag1=0
#initialize the 0 low and 0 colunm
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j]==0:
if i==0:
tag=1
if j==0:
tag1=1
matrix[0][j]=0
matrix[i][0]=0
#fill 0 in the matrix (except the 0 low and 0 colunm)
for i in range(1,len(matrix)):
if matrix[i][0]==0:
for j in range(len(matrix[0])):
matrix[i][j]=0
for j in range(1,len(matrix[0])):
if matrix[0][j]==0:
for i in range(len(matrix)):
matrix[i][j]=0
#The 0 low and 0 colunm
if tag==1:
for j in range(len(matrix[0])):
matrix[0][j] = 0
if tag1==1:
for i in range(len(matrix)):
matrix[i][0] = 0
return matrix
'''