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Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's, and return the matrix.
You must do it in place.
Example 1:
<pre>Input: matrix = [[1,1,1],[1,0,1],[1,1,1]] Output: [[1,0,1],[0,0,0],[1,0,1]] </pre>
Example 2:
<pre>Input: matrix = [[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]] </pre>
Constraints:
m == matrix.lengthn == matrix[0].length1 <= m, n <= 200-2<sup>31</sup> <= matrix[i][j] <= 2<sup>31</sup> - 1
Follow up:
- A straightforward 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?
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题目大意：
若矩阵某元素为0，设置它所在的行和列所有元素均为0，不能用额外区间
解题思路：
用第0行和第0列作为统计。由于第0行和第0列会被覆盖，所以先查看他们有无0
解题步骤：
N/A
注意事项：
- 用第0行和第0列作为统计。由于第0行和第0列会被覆盖，所以先查看他们有无0。两大步骤：先统计，再根据结果设置0
- 第二步中，根据第0和和第0列的结果回设，均从1开始，不含左上cell，因为统计结果不保存在它上。它仅在统计第一行和第一列是否有0时用到。
Python代码：
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23def setZeroes(self, matrix: List[List[int]]) -> None:
# calculate
is_zero_row_zero = True if len([0 for n in matrix[0] if n == 0]) > 0 else False
is_zero_col_zero = True if len([0 for i in range(len(matrix)) if matrix[i][0] == 0]) > 0 else False
for i in range(1, len(matrix)):
for j in range(1, len(matrix[0])):
if matrix[i][j] == 0:
matrix[i][0], matrix[0][j] = 0, 0
# Set
for i in range(1, len(matrix)): # remember to start with 1
if matrix[i][0] == 0:
for j in range(1, len(matrix[0])):
matrix[i][j] = 0
for j in range(1, len(matrix[0])): # remember to start with 1
if matrix[0][j] == 0:
for i in range(1, len(matrix)):
matrix[i][j] = 0
if is_zero_row_zero:
for j in range(len(matrix[0])):
matrix[0][j] = 0
if is_zero_col_zero:
for i in range(len(matrix)):
matrix[i][0] = 0
算法分析：
时间复杂度为<code>O(n<sup>2</sup>)</code>，空间复杂度O(1)
