零矩阵
编写一种算法,若M × N矩阵中某个元素为0,则将其所在的行与列清零。
示例 1:
输入:
[
[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:
[
[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]
]
解题思路:
开辟两个一维数组存放要清理的行列,先遍历一遍,保存需要清理的行列,然后在遍历修改,如果可以使用二维数组也可以先读取原数组需要修改的地方,在新的数组上修改,
java
class Solution {
public void setZeroes(int[][] matrix) {
boolean[] line = new boolean[matrix.length];
boolean[] column = new boolean[matrix[0].length];
// 找出要清零的行列
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] == 0) {
line[i] = true;
column[j] = true;
}
}
}
// 开始对行清零
for (int i = 0; i < matrix.length; i++) {
if (line[i]) {
for (int j = 0; j < matrix[0].length; j++) {
matrix[i][j] = 0;
}
}
}
// 开始对列清零
for (int i = 0; i < matrix[0].length; i++) {
if (column[i]) {
for (int j = 0; j < matrix.length; j++) {
matrix[j][i] = 0;
}
}
}
}
}
python
import copy
class Solution(object):
def setZeroes(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: None Do not return anything, modify matrix in-place instead.
"""
res_x=[]
res_y=[]
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j]==0:
res_x.append(i)
res_y.append(j)
for math in range(len(res_x)):
for x in res_x:
for m in range(len(matrix[0])):
matrix[x][m]=0
for y in res_y:
for n in range(len(matrix)):
matrix[n][y]=0
return matrix
# res=copy.deepcopy(matrix)
# for i in range(len(matrix)):
# for j in range(len(matrix[0])):
# if matrix[i][j]==0:
# for x in range(len(matrix)):
# res[x][j]=0
# print(res[x][j])
# for y in range(len(matrix[0])):
# res[i][y]=0
# return res