题目大意
如果矩阵中存在0,那么把0所在的行和列都置为0。要求在所给的矩阵上完成操作。
注意:最好的空间复杂度是常数空间
解题思路
参考:
https://www.hrwhisper.me/leetcode-set-matrix-zeroes/
https://shenjie1993.gitbooks.io/leetcode-python/073%20Set%20Matrix%20Zeroes.html
思路:
1.创建一个矩阵的拷贝,然后根据这个拷贝进行判断O(MN)
2.创建一个数组,记录矩阵为0的行和列下标O(m+n)
3.把有0的元素映射到首行和首列O(c):详解看代码
代码
这里的复杂度是空间复杂度
O(c)
class Solution(object):
def setZeroes(self, matrix):
""" :type matrix: List[List[int]] :rtype: void Do not return anything, modify matrix in-place instead. """
first_row = False
first_col = False
m = len(matrix)
n = len(matrix[0])
# 第一列若有0,first_col = True
for i in range(m):
if matrix[i][0] == 0:
first_col = True
# 第一行若有0,first_row = True
for j in range(n):
if matrix[0][j] == 0:
first_row = True
# 若中间有0,同时映射到首行首列
for i in range(1, m):
for j in range(1, n):
if matrix[i][j] == 0:
matrix[i][0] = matrix[0][j] = 0
# 然后通过首行首列,将所有改为0的置为0
for i in range(1, m):
for j in range(1, n):
if matrix[0][j] == 0 or matrix[i][0] == 0:
matrix[i][j] = 0
# 最后再把保存好的首行首列是否改为0的信息拿出来置为0
if first_row:
for j in range(n):
matrix[0][j] = 0
if first_col:
for i in range(m):
matrix[i][0] = 0
O(m+n)
class Solution:
# @param matrix, a list of lists of integers
# RETURN NOTHING, MODIFY matrix IN PLACE.
def setZeroes(self, matrix):
m , n = len(matrix), len(matrix[0])
row , col = [0 for i in range(m)] , [0 for i in range(n)]
for i in range(m):
for j in range(n):
if not matrix[i][j]:
row[i]=col[j]=1
for i in range(m):
if row[i]:
for j in range(n):
matrix[i][j]=0
for j in range(n):
if col[j]:
for i in range(m):
matrix[i][j]=0
O(mn)
class Solution:
# @param matrix, a list of lists of integers
# RETURN NOTHING, MODIFY matrix IN PLACE.
def setZeroes(self, matrix):
m , n = len(matrix), len(matrix[0])
temp = [[matrix[i][j] for j in range(n)] for i in range(m)]
for i in range(m):
for j in range(n):
if not temp[i][j]:
self.setZero(i,j,n,m,matrix)
def setZero(self,row,col,n,m,matrix):
for i in range(m):
matrix[i][col]=0
for j in range(n):
matrix[row][j]=0