## 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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
[1,   4,  7, 11, 15],
[2,   5,  8, 12, 19],
[3,   6,  9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]


Given target = 5, return true.

Given target = 20, return false.

## Solutions

右上角向下向左遍历。

# Time: O(nlogn)
# Space: O(nlogn)
class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
if not matrix or not matrix[0]:
return False

r = len(matrix)
for i in range(r):
if matrix[i][-1] < target:
continue
if self.binary_search(matrix[i], target):
return True
return False

def binary_search(self, nums, target):
if not nums:
return False
l, r = 0, len(nums) - 1
while l <= r:
mid = l + (r - l) // 2
if nums[mid] == target:
return True
elif nums[mid] > target:
r = mid - 1
else:
l = mid + 1
return False

# 129/129 cases passed (32 ms)
# Your runtime beats 82.94 % of python3 submissions
# Your memory usage beats 88.89 % of python3 submissions (17.4 MB)


### 2. Trick

# Time: O(m+n)
# Space: O(1)
class Solution:
def searchMatrix(self, matrix, target):
if not matrix or not matrix[0]:
return False

r, c = 0, len(matrix[0]) - 1
while r < len(matrix) and c >= 0:
if matrix[r][c] == target:
return True
elif matrix[r][c] > target:
c -= 1
else:
r += 1
return False

# 129/129 cases passed (40 ms)
# Your runtime beats 36.96 % of python3 submissions
# Your memory usage beats 92.59 % of python3 submissions (17.4 MB)