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.

假设直接对矩阵元素进行二分查找的话，时间复杂度是O(m*n)，事实上非常easy想到先通过查找找到相应可能存在于哪一行，然后再在那行中查找是否存在，採用最简单的直接查找这样时间复杂度仅有O(m+n)，假设这两次查找再分别採用二分查找的话，时间复杂度更能够减少到O（logm+logn），以下是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;

    }

};