Leetcode:search a 2D Matrix

Search a 2D Matrix

Total Accepted: 43629 submissions: 138231

Write an efficient algorithm, searches for a value in a m x n Matrix. This matrix has the following properties:

• Integers in each row is sorted from the left to the right.
• The first integer of each row was greater than the last integer of the previous row.

For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [Ten, One,],  [23, 30, 34, 50]]

Given Target = 3 , return true .

Conventional train of thought, two times

BOOL Searchmatrix (vector<vector<int>>& matrix, int target) {        if (Matrix.empty ())            return False ;        Vector<int> Firstcol;        for (auto V:matrix)        firstcol.push_back (v[0]);        if (Binary_search (Firstcol.begin (), Firstcol.end (), target))            return true;         Vector<int>::iterator iter = Lower_bound (Firstcol.begin (), Firstcol.end (), target);        int low = Iter-firstcol.begin ();        for (int i = 0; i < low; i++)        {            if (Binary_search (Matrix[i].begin (), Matrix[i].end (), target))                return true;        }        return false;    }

Opportunistic, there are potential bugs

BOOL Searchmatrix (vector<vector<int> >& matrix, int target) {        int rows = Matrix.size ();        if (rows = = 0)    return false;        int cols = Matrix[0].size ();        int first = 0, last = rows * COLS-1;        if (Target < matrix[0][0] | | target > MATRIX[ROWS-1][COLS-1])            return false;        int mid, Val;        Binary search        while (first <=) {            mid = first + (Last-first)/2;            val = matrix[mid/cols][mid% cols];            if (Val > Target) Last                = mid-1;            else if (val = = target)                return true;            else First                = mid + 1;        }        return false;    }

Leetcode:search a 2D Matrix