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.

这个题目实际上考查的是二分查找，通过下标操作可以将矩阵变成一个一维升序数组，我的C++实现代码如下：

bool searchMatrix(vector<vector<int> > &matrix, int target) {
    int row = matrix.size(), col = matrix[0].size();

    int low = 0, high = row * col - 1;
    while (low <= high) {
        int middle = (low + high) >> 1;
        int midVal = matrix[middle / col][middle % col];
        if (midVal == target) return true;
        else if (midVal > target) high = middle - 1;
        else low = middle + 1;
    }

    return false;
}

LeetCode[Array]: Search a 2D Matrix