動機
這種題目是不是就是saddleback就好了
Problem
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],[23,30,34,60]], target = 3Output: true
Example 2:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13Output: false
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-104 <= matrix[i][j], target <= 104
Sol
class Solution:
def searchMatrix(self, mx: List[List[int]], target: int) -> bool:
a,b = len(mx)-1, 0
while 0 <= a < len(mx) and 0 <= b < len(mx[a]):
if mx[a][b] == target:
return True
elif mx[a][b] > target:
a -= 1
else:
b += 1
return False