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Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

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• 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.

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  For example,

  Consider the following matrix:

  [
    [1,   3,  5,  7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
  ]
  

  Given target = 3, return true.

  
  

  由于这个矩阵是有序的，我们可以从右上角开始进行比较，若比右上角大，则肯定在下一行，否则就在本行中，时间复杂度O（m+n）

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  当然也可以用二分来做，同样从最右边那一列开始看，先二分出是在哪一行，然后再二分出是在哪一列时间复杂度 O(logn+logm)

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  下面给出O(m+n)的代码

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

  
  
  

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