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.

Analyse: binary search. 

Runtime: 12ms.

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.empty() || matrix[0].empty()) return false;
        
        int m = matrix.size(), n = matrix[0].size();
        // target the row, low would be the target
        int low = 0, high = m - 1;
        while(low < high) {
            int mid = low + (high - low) / 2;
            if(matrix[mid][n - 1] > target) high = mid;
            else if(matrix[mid][n - 1] < target) low = mid + 1;
            else return true;
        }
        
        // target the column
        int start = 0, end = n - 1;
        while(start < end) {
            int mid = start + (end - start) / 2;
            if(matrix[low][mid] > target) end = mid;
            else if(matrix[low][mid] < target) start = mid + 1;
            else return true;
        }
        return matrix[low][start] == target;
    }
};