Search a 2D Matrix

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.

  
  

  
  题意：在一个二维矩阵中找到给定的值。矩阵从上到下从左到右有序
  思路：二维空间的二分查找
  先在一维里找中间位置，再将该位置转为二维空间里的下标
  复杂度：
  
  

  class Solution {
  public:
      bool searchMatrix(const vector
      
        > &matrix, int target){
      	if (matrix.empty()) return false;
      	int m = matrix.size(), n = matrix.front().size();
      	int begin = 0, end = m * n, middle, row, col;
  
  
      	while(begin < end){
      		middle = begin + (end - begin) / 2;
      		int row = middle / n;
      		int col = middle % n;
      		if(matrix[row][col] == target) return true;
      		else if(matrix[row][col] < target) begin = middle + 1;
      		else end = middle;
      	}
      	return false;
      }
  };