二分查找经典题目；根据矩阵的特点，不需要把矩阵拉成一维，二维转成一维映射关系为a[i]=matrix[⌊i//n⌋][i%n]；然后开始二分查找，一直二分的收缩区间；
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
)
