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.

解题分析：这道题很简单，也没什么难度，就是一个普通的二分搜索的应用，可能稍微难的地方，就是二维矩阵位置的转换。

int  middle = (low + high) / 2;
if(matrix[middle / n][middle % n] == target)

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int m = matrix.size();
        int n = matrix[0].size();

        int low = 0;
        int high = m*n-1;
        while(low <= high)
        {
             int  middle = (low + high) / 2;
            if(matrix[middle / n][middle % n] == target)
                return true;
            else if(matrix[middle / n][middle % n] > target)
                 high = middle -1;
            else if(matrix[middle / n][middle % n] < target)
                 low = middle -1;
        }
        return false;


    }
};

LeetCode-Search a 2D Matrix