LeetCode Ts-766. Toeplitz Matrix

    AllenLi
    CodingLeetCode Js-Matrix

    Given an m x n matrix, return true if the matrix is Toeplitz. Otherwise, return false.
    A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same elements.

    給予一個 m x n 的矩陣，如果矩陣中「每條從左上到右下的對角線」上的元素都相同，則回傳 true，否則回傳 false。

    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.

    Solution:
    這題要能理解如何從二維陣列中取出數值來進行比對，
    再來就是遍歷矩陣中的陣列元素，
    如果左上與右下相同則繼續直到結束回傳 true，否則回傳 false。

    Code 1: BigO(m * n)

    function isToeplitzMatrix(matrix: number[][]): boolean { 
    for (let i = 1; i < matrix.length; i++) {
        for (let j = 1; j < matrix[i].length; j++) {
            if (matrix[i][j] !== matrix[i - 1][j - 1]) {
                return false;
            }
        }
    }
    return true;
};

    FlowChart:
    Example 1

    Input: matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]]

matrix[i][j] 1, matrix[i - 1][j - 1] 1 // continue;
matrix[i][j] 2, matrix[i - 1][j - 1] 2 // continue;
matrix[i][j] 3, matrix[i - 1][j - 1] 3 // continue;
matrix[i][j] 5, matrix[i - 1][j - 1] 5 // continue;
matrix[i][j] 1, matrix[i - 1][j - 1] 1 // continue;
matrix[i][j] 2, matrix[i - 1][j - 1] 2 // continue;

return true;

    Example 2

    Input: matrix = [[1,2],[2,2]]

matrix[i][j] 2, matrix[i - 1][j - 1] 1 // false

return false;

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