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766. Toeplitz Matrix
The most simple version: class Solution { public boolean isToeplitzMatrix(int[][] matrix) { if(matrix==null||matrix[0].length==0) return true; int m = matrix.length, n = matrix[0].length; for(int i=0;i<m-1;i++){ for(int j=0;j&766. Toeplitz Matrix
Don't think about a lot, this is a very easy problem, time complexity is O(m*n) public boolean isToeplitzMatrix(int[][] matrix) { if (matrix == null || matrix[0].length == 0) return true; int m = matrix.length, n = ma766. Toeplitz 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. Example 1: Input: matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]] Output: trToeplitz定理推广和应用
Toeplitz定理推广和应用 Toeplitz定理 设 n,k∈Nn,k \in \mathbb{N}n,k∈N，tnk≥0t_{nk}\ge0tnk≥0 且 ∑k=1ntnk=1\sum_{k=1}^{n}t_{nk}=1∑k=1ntnk=1，limn→+∞tnk=0\lim_{n\rightarrow +\infty}t_{nk}=0limn→+∞tnk=0。如果 limn→+∞an=a\lim_{n \rightarr【LEETCODE】45、766. Toeplitz Matrix
package y2019.Algorithm.array;/** * @ProjectName: cutter-point * @Package: y2019.Algorithm.array * @ClassName: IsToeplitzMatrix * @Author: xiaof * @Description: 766. Toeplitz Matrix * A matrix is Toeplitz if every diagonal from top-left to bottom-right ha算法——托普利茨矩阵
托普利茨矩阵 # ToeplitzMatrix 托普利茨矩阵class Solution: def is_toeplitz_matrix(self, matrix): for i in range(len(matrix[0])): for v in range(1, len(matrix)): try: if matrix[0][i] != matrix[v][i+v]: