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.
For 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.

Python

# time O(m) + O(n)
# matrix[:,end]< target,則該row所有元素均小于target
# matrix[:,end] > target, 則搜索該row 可能發(fā)現(xiàn)target
# 從右上角開始搜
class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if matrix:
            m = len(matrix)
            n = len(matrix[0])
            r = 0
            c  = n - 1 
            while r < m and c >= 0:
                if matrix[r][c] == target:
                    return True
                if matrix[r][c] > target:
                    c -= 1
                else:
                    r += 1
        return False