Jianzhi offer brushing notes-find a two-dimensional array

The idea I think is to compare the size of the target and the center value of the matrix. If the target is large, the 1/4 matrix in the upper left of the center is excluded, and if the target is small, the 1/4 matrix in the lower right is excluded. Then divide the remaining 3/4 matrix into three parts, recursively call the function to find the three matrices.
code show as below:

import numpy as np
class Solution:
    # array 二维列表
    def Find(self, target, array):
        # write code here
        print(array)
        array = np.array(array)
        if array.size == 0:
            return False
        col_r = array.shape[0]
        row_r = array.shape[1]
        col_m = int((col_r-1)/2)
        row_m = int((row_r-1)/2)
        anchor = array[col_m, row_m]
        if target < anchor:
            return self.Find(target, array[:col_m,:row_m]) or\
             self.Find(target, array[col_m:col_r,:row_m]) or\
              self.Find(target, array[:col_m,row_m:row_r])
        elif target > anchor:
            return self.Find(target, array[col_m+1:col_r,row_m+1:row_r]) or\
             self.Find(target, array[col_m+1:col_r,:row_m+1]) or\
              self.Find(target, array[:col_m+1,row_m+1:row_r])
        else:
            return True

However, the test speed is very low, because it is the complexity of log(4/3, m n). When the matrix is ​​larger than 4, the speed will be slower than O(m+n). So use binary search to search from the upper right corner, the code is as follows:

class Solution(object):
    def findNumberIn2DArray(self, matrix, target):
        if len(matrix) == 0:
            return False
        r = len(matrix[0])-1
        t = 0
        max_c = len(matrix)
        while r >= 0 and t < max_c:
            anchor = matrix[t][r]
            if target == anchor:
                return True
            elif target > anchor:
                t+=1
            else:
                r-=1
        return False

This question is still relatively simple, that is, the code is not skilled enough, pay attention to the judgment of the empty input matrix