## Description

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input:
matrix = [
[1,   3,  5,  7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true


Example 2:

Input:
matrix = [
[1,   3,  5,  7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false


## Solutions

要找到trick，每一行最右边的数都比该行前面的数大，那么先可以通过每一行最后一个数确定行，然后采用 binary search 即可：

# Time: O(nlogm)
# Space: O(n)
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
if not matrix or len(matrix) == 0:
return False
h, w = len(matrix), len(matrix)
r = -1
for i in range(h):
if w-1 >= 0 and matrix[i][w-1] >= target:
r = i
break

if r == -1:
return False
if self.binary_search(matrix[r], target) == -1:
return False
else:
return True

def binary_search(self, nums, target):
if not nums:
return -1

n = len(nums)
l, r = 0, n - 1
while l <= r:
mid = (l + r) >> 1
if nums[mid] == target:
return mid
elif nums[mid] > target:
r = mid - 1
else:
l = mid + 1
return -1
# 136/136 cases passed (68 ms)
# Your runtime beats 92.86 % of python3 submissions
# Your memory usage beats 5.88 % of python3 submissions (14.9 MB)