二分查找經典題目；根據矩陣的特點，不需要把矩陣拉成一維，二維轉成一維映射關系為a[i]=matrix[?i//n?][i%n]；然后開始二分查找，一直二分的收縮區間；
class Solution:def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:m = len(matrix)n = len(matrix[0])left, right = -1, m*nwhile left+1 < right:mid = (left+right)//2x = matrix[mid//n][mid%n]if x == target:return Trueif x < target:left = midelse:right = midreturn False
)
