Problem description:

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.
Example 1:

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Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

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Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

Solution:

The idea is to think the matrix as an array.
array convert to n * m matrix => a[x] => matrix[x / m][x % m];

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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();
        int left= 0, right= m*n-1;
        
        while(left <= right){
            int mid= left+ (right-left)/2;
            int tmp= matrix[mid/n][mid%n];
            if(target> tmp)
                left= mid+1;
            else if(target< tmp)
                right= mid-1;
            else
                return true;
        }
        return false;
    }
};

time complexity: $O(log(mn))$
space complexity: $O(1)$
reference:
https://goo.gl/FAmm72