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.

如果直接对矩阵元素进行二分查找的话，时间复杂度是O(m*n)，其实很容易想到先通过查找找到对应可能存在于哪一行，然后再在那行中查找是否存在，采用最简单的直接查找这样时间复杂度仅有O(m+n)，如果这两次查找再分别采用二分查找的话，时间复杂度更可以降低到O（logm+logn），下面是O（m+n）的代码：

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if(matrix.empty())
            return false;
        int m = matrix.size();
        int n = matrix[0].size();
        int i = 0, j=0;
        while(i<m && target>=matrix[i][0])
            i++;
        i--;
        if(i==-1)
            return false;
        while(j<n)
        {
            if(target == matrix[i][j])
                return true;
            else
                j++;
        }
        return false;
    }
};