[Leetcode]#74 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.

//#74 Search a 2D Matrix
//16ms 45.83%
class Solution {
public:
bool searchMatrix(vector <vector<int> > & matrix, int target)
{
vector<int> m;
for(unsigned int i=0; i<matrix.size(); i++)
{
for(unsigned int j=0; j<matrix[i].size(); j++)
{
m.push_back(matrix[i][j]);
}
}
int ub(matrix.size() * matrix[0].size());
int lb(0);

while(lb != ub)
{
//cout << lb << ", " << ub << endl;
int p = (lb+ub)/2;
if(m[p] == target)
{
return true;
}
else if(lb == ub - 1)
{
return false;
}
if(m[p] > target)
{
//cout << "Too Large\n";
ub = p;
}
else
{
//cout << "Too Small\n";
lb = p;
}
}

return false;
}
};