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原题
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
Example 1:
Input:
[
[1,1,1],
[1,0,1],
[1,1,1]
]
Output:
[
[1,0,1],
[0,0,0],
[1,0,1]
]
Example 2:
Input:
[
[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]
]
Follow up:
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?
解法1
首先将原来的matrix深度复制给prev, 遍历prev, 如果找到0, 则将它所在的行和列都变为0.
Time: O(mnm*n)
Space: O(1)
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.
"""
# edge case
if not matrix:
return
prev = copy.deepcopy(matrix)
row, col = len(prev), len(prev[0])
for i in range(row):
for j in range(col):
if prev[i][j] == 0:
# set the entire row and col to 0
matrix[i] = [0]*col
for r in range(row):
matrix[r][j] = 0
解法2
遍历matrix, 记录0所在的行和列, 然后对行和列进行赋值.
Time: O(m*n)
Space: O(1)
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.
"""
# edge case
if not matrix:
return
row, col = len(matrix), len(matrix[0])
r_list, c_list = [], []
for i in range(row):
for j in range(col):
if matrix[i][j] == 0:
r_list.append(i)
c_list.append(j)
# change the target rows and cols to 0
for r in r_list:
matrix[r] = [0]*col
for c in c_list:
for r in range(row):
matrix[r][c] = 0