562. Longest Line of Consecutive One in Matrix

This is an easy question.  No matter what kind of method you chose, the time complexity is at least O(m*n), becasue you have to traverse each element at least once. Here, we have a straight forward method without any maps...

class Solution {
public:
    int longestLine(vector<vector<int>>& M) {
        if (M.empty()) return 0;
        int result = 0;
        int cur = 0;
        for (int i = 0; i < M.size(); i++) {
            cur = 0;
            for (int j = 0; j < M[i].size(); j++) {
                if (M[i][j]) cur++;
                else cur = 0;
                result = max(result, cur);
            }
        }
        for (int j = 0; j < M[0].size(); j++) {
            cur = 0;
            for (int i = 0; i < M.size(); i++) {
                if (M[i][j]) cur++;
                else cur = 0;
                result = max(result, cur);
            }
        }

        //int my = M.size(), mx = M[0].size();
        for (int j = -(int)M.size(); j < (int)M[0].size(); j++) {
            cur = 0;
            for (int i = 0; i < M.size(); i++) {
                if (j + i < 0 || i + j >= M[0].size()) continue;
                else {
                    if (M[i][j + i]) cur++;
                    else cur = 0;
                    result = max(result, cur);
                }
            }
        }
        for (int j = 0; j < M[0].size() + M.size(); j++) {
            cur = 0;
            for (int i = 0; i < M.size(); i++) {
                if (j - i < 0 || j - i >= M[0].size()) continue;
                else {
                    if (M[i][j - i]) cur++;
                    else cur = 0;
                    result = max(result, cur);
                }
            }
        }
        return result;
    }
};

Pay attention to the third for-loop. Because the return value of size() is size_type, an unsigned value, when you make some operations with int, be very careful about bugs, especially when the value is minus.