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.

Solution：二分查找

思路：将矩阵按顺序直接当成递增序列的二分查找来做，不同的是对index元素的访问需要转换成二维数组的row和col。
Time Complexity: O(N) Space Complexity: O(1)
N为元素个数

Solution Code：

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0) {
            return false;
        }
        int n = matrix.length, m = matrix[0].length;
        int start = 0, end = m * n - 1;
        while(start <= end) {
            int mid = (start + end) / 2;
            int value = matrix[mid / m][mid % m];
            
            if(value == target) return true;
            else if(value < target) {
                start = mid + 1;
            }
            else {
                end = mid - 1;
            }
        }
        return false;
    }
}