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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