#!/usr/bin/env python

# encoding: utf-8

"""

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].

Follow up:

What if the matrix is stored on disk, and the memory is limited such that you can only load at most one row of the matrix into the memory at once?

What if the matrix is so large that you can only load up a partial row into the memory at once?

"""

class Solution:

def isToeplitzMatrix(self, matrix):

"""

:type matrix: List[List[int]]

:rtype: bool

>>> matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]]

>>> s = Solution()

>>> s.isToeplitzMatrix(matrix)

True

"""

# 两层遍历

row = len(matrix)

col = len(matrix[0])

for i in range(row - 1):

for j in range(col - 1):

if matrix[i][j] != matrix[i + 1][j + 1]:

return False

return True

# 找数据规律，每层跟下一层如果对角线值一致，会部分相等的。

# return all(matrix[row+1][1:] == matrix[row][:-1] for row in range(len(matrix)-1))

if __name__ == '__main__':

import doctest

doctest.testmod(verbose=True)