LintCode "Find Peak Element II"

Idea is the same: climbing up the hill along one edge (Greedy)! Visualize it in your mind!

class Solution {
public:
    /**
     * @param A: An integer matrix
     * @return: The index of the peak
     */
    vector<int> findPeakII(vector<vector<int> > A)
    {
        int n = A.size();
        int m = A[].size();

        int i = , j = ;
        while(i < m -  && j < n - )
        {
            bool up   = A[j - ][i] < A[j][i];
            bool down = A[j + ][i] < A[j][i];
            bool left = A[j][i - ] < A[j][i];
            bool right= A[j][i + ] < A[j][i];

            if(up && down && left && right)
            {
                return {j, i};
            }
            if(!down && j < n - )
            {
                j ++;
            }
            else if(!right && i < m - )
            {
                i ++;
            }
        }

        return {-, -};
    }
};