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给定一个 m x n
的矩阵,如果一个元素为 0 ,则将其所在行和列的所有元素都设为 0 。请使用 原地(http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95) 算法。
进阶:
O(mn)
的额外空间,但这并不是一个好的解决方案。O(m + n)
的额外空间,但这仍然不是最好的解决方案。示例 1:
输入:matrix = [[1,1,1],[1,0,1],[1,1,1]] 输出:[[1,0,1],[0,0,0],[1,0,1]]
示例 2:
输入:matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]] 输出:[[0,0,0,0],[0,4,5,0],[0,3,1,0]]
提示:
m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-2^31 <= matrix[i][j] <= 2^31 - 1
- class Solution(object):
- def setZeroes(self, matrix):
- """
- :type matrix: List[List[int]]
- :rtype: void Do not return anything, modify matrix in-place instead.
- """
- if not matrix:
- return
- m = len(matrix)
- if m == 0:
- return
- r = []
- c = []
- n = len(matrix[0])
- for i in range(m):
- for j in range(n):
- if matrix[i][j] == 0:
- r.append(i)
- c.append(j)
- r = set(r)
- c = set(c)
- for i in r:
- for j in range(n):
- matrix[i][j] = 0
- for i in range(m):
- for j in c:
- matrix[i][j] = 0
- return matrix
- # %%
- s = Solution()
- print(s.setZeroes(matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]))
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