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
      
        > &matrix, int target) {
          if(matrix.empty())
              return false;
          int m = matrix.size();
          int n = matrix[0].size();
          int i = 0, j=0;
          while(i
       
        =matrix[i][0]) i++; i--; if(i==-1) return false; while(j