leetcode讲解--766. Toeplitz Matrix

题目

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:

1. 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?
2. What if the matrix is so large that you can only load up a partial row into the memory at once?

讲解

这道题我一开始在while里面只用了一个限制条件，后来发现不行的。其实多加一些可能会出现的条件是一种很好的习惯。

Java代码

class Solution {
    public boolean isToeplitzMatrix(int[][] matrix) {
        for(int i=0;i<matrix.length;i++){
            int row=i;
            int column=0;
            while(row<matrix.length-1 && column<matrix[0].length-1){
                if(matrix[row][column]!=matrix[row+1][column+1]){
                    return false;
                }
                column++;
                row++;
            }
        }
        for(int i=0;i<matrix[0].length;i++){
            int column=i;
            int row=0;
            while(row<matrix.length-1 && column<matrix[0].length-1){
                if(matrix[row][column]!=matrix[row+1][column+1]){
                    return false;
                }
                column++;
                row++;
            }
        }
        return true;
    }
}