LeetCode 073 Set Matrix Zeroes

LeetCode

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:

Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[1,0,1],[0,0,0],[1,0,1]]

Example 2:

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]]

Constraints:

m == matrix.length n == matrix[0].length
1 <= 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?

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：

1. 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
2. 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：

def 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O(n2)，空间复杂度O(1)