74. Search a 2D Matrix

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

解题要点：

把整个2d数组看作一条1d数组，长度可以计算为n * m - 1。然后用二分算法搜素目标数字。

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        n = len(matrix)
        if n == 0:
            return False
        m = len(matrix[0])
        if m == 0:
            return False
        l = 0
        h = m * n - 1
        while l <= h:
            mid = l + (h-l) / 2
            i = mid / m
            j = mid % m
            if matrix[i][j] == target:
                return True
            elif matrix[i][j] < target:
                l = mid + 1
            else:
                h = mid - 1
        return False