给定一个N×N的矩阵，0和1。设置一个包含每一行 0 所有 0 和设置包含每列 0 所有 0 秒。

Given a NxN matrix with 0s and 1s. Set every row that contains a 0 to all 0s and set every column that contains a 0 to all 0s.

例如

1 0 1 1 0
0 1 1 1 0
1 1 1 1 1
1 0 1 1 1
1 1 1 1 1

结果

0 0 0 0 0
0 0 0 0 0
0 0 1 1 0
0 0 0 0 0
0 0 1 1 0

一个微软工程师告诉我，有一个解决方案，不涉及额外的内存，只有两个布尔变量和一个传球，所以我在寻找这个问题的答案。

A Microsoft Engineer told me that there is a solution that involves no extra memory, just two boolean variables and one pass, so I'm looking for that answer.

顺便说一句，想象它是一个有点矩阵，因此只需1和0被允许在基质中。

BTW, imagine it is a bit matrix, therefore just 1s and 0s are allow to be in the matrix.

推荐答案

好了，我也累了，因为它是上午03点在这里，但我有一个第一次尝试就地恰好有2次通过矩阵中的每个数字，因此为O （N×N个），它是线性的矩阵的大小。

Ok, so I'm tired as it's 3AM here, but I have a first try inplace with exactly 2 passes on each number in the matrix, so in O(NxN) and it is linear in the size of the matrix.

我用1rst列和第一行作为标记知道哪里是行/ COLS只有1的。然后，有2个变量L和C要记住，如果1rst行/列全部为1也。 所以第一遍设置标记，其余的重置为0。

I use 1rst column and first row as markers to know where are rows/cols with only 1's. Then, there are 2 variables l and c to remember if 1rst row/column are all 1's also. So the first pass sets the markers and resets the rest to 0's.

第二遍集1中的地方被标行列数为1，并复位第一行/列取决于升和c。

The second pass sets 1 in places where rows and cols where marked to be 1, and resets 1st line/col depending on l and c.

我怀疑坚信我一定能在1通来完成在年初广场取决于到底广场。也许我的第2次，可以更加高效......

I doubt strongly that I can be done in 1 pass as squares in the beginning depend on squares in the end. Maybe my 2nd pass can be made more efficient...

import pprint

m = [[1, 0, 1, 1, 0],
     [0, 1, 1, 1, 0],
     [1, 1, 1, 1, 1],
     [1, 0, 1, 1, 1],
     [1, 1, 1, 1, 1]]



N = len(m)

### pass 1

# 1 rst line/column
c = 1
for i in range(N):
    c &= m[i][0]

l = 1
for i in range(1,N):
    l &= m[0][i]


# other line/cols
# use line1, col1 to keep only those with 1
for i in range(1,N):
    for j in range(1,N):
        if m[i][j] == 0:
            m[0][j] = 0
            m[i][0] = 0
        else:
            m[i][j] = 0

### pass 2

# if line1 and col1 are ones: it is 1
for i in range(1,N):
    for j in range(1,N):
        if m[i][0] & m[0][j]:
            m[i][j] = 1

# 1rst row and col: reset if 0
if l == 0:
    for i in range(N):
        m [i][0] = 0

if c == 0:
    for j in range(1,N):
        m [0][j] = 0


pprint.pprint(m)