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.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

注意这道题给出的矩阵，每行都是有序数组，并且每一行的结尾数字是小于下一行的开头数字的。注意这道题与剑指offer上的《面试题3：二维数组中的查找》给出的二维数组的特性是不一样的。这道题有明显的使用二分搜索的特点。

一种做法是我一开始想到，先二分到行，确定一行用于查找，之后再在行里查找，感觉我还是没有把二分越边界的情况想的特别清楚，代码有些冗长，时间复杂度是O(logm+logn),空间复杂度O(1),代码如下：

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or not matrix[0] or target < matrix[0][0] or target > matrix[-1][-1]:
            return False 
        m = len(matrix)
        n = len(matrix[0])
        
        l = 0
        r = m-1
        
        while l < r:
            mid = l + (r-l)/2
            if target > matrix[mid][-1]:
                l = mid+1
            elif target < matrix[mid][0]:
                r = mid-1
            else:
                l = r = mid
                break
        if l != r:
            return False
            

        if target < matrix[l][0] or target > matrix[l][-1]:
            return False
        k = l
        l = 0
        r = n-1
        while l <= r:
            mid = l+(r-l)/2
            if target < matrix[k][mid]:
                r = mid -1 
            elif target > matrix[k][mid]:
                l = mid +1
            else:
                return True
        return False

另外一种简洁直接的解法为将二维数组看作一维排序数组来考虑，唯一需要注意的是如何将mid的值转化为实际二维数组中的index，注意是除以行宽呀，盆友。时间复杂度O(log(mn))=O(log(m)+log(n)),代码如下，简洁许多：

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or not matrix[0]:
            return False
        m = len(matrix)
        n = len(matrix[0])
        
        l = 0
        r = m*n -1
        while l <= r:
            mid = l + (r-l)/2
            if target < matrix[mid/n][mid%n]:
                r = mid -1
            elif target > matrix[mid/n][mid%n]:
                l = mid +1
            else:
                return True
        return False