Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
Follow up:
Did you use extra space?
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?
Set Matrix Zeroes
题目本身很简单，最简单的方式是记录下每个有0的row和column，然后通过go through 一遍就可以，while ，如果所有的element都是0的话，那space就是m*n
但是可以通过一种很巧妙的方法规避掉额外空间
因为第一个row和column终归会变成0如果他所在的column or row有0存在，那我们为什么不用first row 和first column来记录呢
| 0 | 1 | 1 | 0 | 1 |
|---|---|---|---|---|
| 1 | ||||
| 0 | ||||
| 1 |
每当我们看到一个0的时候，我们就可以martix[i][0] = 0, matrix[0][j] = 0
后面的时候我们只需要每次查看element当前的row or column是否是 0，如果是的话我们就martix[i][j] = 0,
具体实现看代码
class Solution(object):
def setZeroes(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: void Do not return anything, modify matrix in-place instead.
"""
r ,c = 1,1
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == 0:
matrix[i][0] = 0
matrix[0][j] = 0
if i == 0:
r = 0
if j == 0:
c = 0
for i in range(len(matrix)):
for j in range(len(matrix[0])):
print(matrix)
if(matrix[i][0] == 0 or matrix[0][j] == 0):
matrix[i][j] = 0
if not r:
matrix[0] = [0]*len(matrix[0])
if not c:
for i in range(len(matrix)):
matrix[i][0] = 0
