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.

  这是一个杨氏矩阵，即每一行是递增的，每一列也是递增的。根据递增性质，我们对每一行只搜索最后一个数，如果小于target那么前往下一行，如果大于target那么说明在这一行，于是前往前一列，直到找到该数。

  ?

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

  

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  这道题难度是medium，本来我打算先把easy难度的做完在考虑中等难度的。偶然看了七月算法的视频（排序查找实战），受益匪浅，决定对讲过的例题进行巩固消化，于是先把这道题做了，算法思路来源于曹博，学习了！
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