![[LeetCode&Python] Problem 766. Toeplitz Matrix [LeetCode&Python] Problem 766. Toeplitz Matrix](https://image.shishitao.com:8440/aHR0cHM6Ly9ia3FzaW1nLmlrYWZhbi5jb20vdXBsb2FkL2NoYXRncHQtcy5wbmc%2FIQ%3D%3D.png?!?w=700&webp=1)
A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same element.
Now given an M x N
matrix, return True
if and only if the matrix is Toeplitz.
Example 1:
Input:
matrix = [
[1,2,3,4],
[5,1,2,3],
[9,5,1,2]
]
Output: True
Explanation:
In the above grid, the diagonals are:
"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]".
In each diagonal all elements are the same, so the answer is True.
Example 2:
Input:
matrix = [
[1,2],
[2,2]
]
Output: False
Explanation:
The diagonal "[1, 2]" has different elements.
Note:
-
matrix
will be a 2D array of integers. -
matrix
will have a number of rows and columns in range[1, 20]
. -
matrix[i][j]
will be integers in range[0, 99]
.
class Solution:
def isToeplitzMatrix(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: bool
"""
Flag=True
m=len(matrix)
n=len(matrix[0]) for i in range(m):
row=i
column=0
stan=matrix[i][0]
while row<m and column<n:
if matrix[row][column]!=stan:
Flag=False
break
row+=1
column+=1 if Flag:
for j in range(1,n):
column=j
row=0
stan=matrix[0][j]
while column<n and row<m:
if matrix[row][column]!=stan:
Flag=False
break
row+=1
column+=1 return Flag