[leetcode] 766. Toeplitz Matrix @ python

原题

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?

解法1

先从左到右, 再从上到下遍历matrix的对角线, 将对角线的元素放入集合里, 检查集合的长度是否大于1.

代码

class Solution(object):def isToeplitzMatrix(self, matrix):""":type matrix: List[List[int]]:rtype: bool"""row, col = len(matrix), len(matrix[0])# check from left to rightfor y in range(col):diag  = set()x = 0diag.add(matrix[x][y])while x+1 < row and y+1 < col:x += 1y += 1diag.add(matrix[x][y])# checkif len(diag) > 1:return False# check from up to downfor x in range(1, row):diag = set()y = 0diag.add(matrix[x][y])while x+1 < row and y+1 < col:x += 1y += 1diag.add(matrix[x][y])# checkif len(diag) > 1:return Falsereturn True

解法2

直接遍历matrix, 如果发现某个元素与它的对角线元素不相等, 返回False.

代码

class Solution(object):def isToeplitzMatrix(self, matrix):""":type matrix: List[List[int]]:rtype: bool"""row, col = len(matrix), 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 Falsereturn True

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[leetcode] 766. Toeplitz Matrix @ python

原题

解法1

代码

解法2

代码

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