Longest Line of Consecutive One in Matrix

Given a 01 matrix M, find the longest line of consecutive one in the matrix. The line could be horizontal, vertical, diagonal or anti-diagonal.

Example:
Input:
[[0,1,1,0],
[0,1,1,0],
[0,0,0,1]]
Output: 3

 1 public class Solution {
 2         int[][] directions = new int[][] { { 1, 0 }, { 0, 1 }, { 1, 1 }, { 1, -1 } };
 3 
 4         public int longestLine(int[][] M) {
 5             if (M == null || M.length == 0 || M[0].length == 0)
 6                 return 0;
 7             int res = 0;
 8             for (int i = 0; i < M.length; i++) {
 9                 for (int j = 0; j < M[0].length; j++) {
10                     if (M[i][j] == 1) {
11                         res = Math.max(res, getMax(M, i, j));
12                     }
13                 }
14             }
15             return res;
16         }
17 
18         private int getMax(int[][] M, int x, int y) {
19             int res = 1;
20             for (int[] dir : directions) {
21                 int i = x + dir[0];
22                 int j = y + dir[1];
23                 int count = 1;
24                 while (i < M.length && j < M[0].length && i >= 0 && j >= 0 && M[i][j] == 1) {
25                     i += dir[0];
26                     j += dir[1];
27                     count++;
28                 }
29                 res = Math.max(count, res);
30             }
31             return res;
32         }
33         
34         int longestLineV2(int[][] M) {
35             if (M == null || M.length == 0 || M[0].length == 0) return 0;
36             int m = M.length, n = M[0].length, res = 0;
37             int[][] dirs = { { 1, 0 }, { 0, 1 }, { -1, -1 }, { -1, 1 } };
38             for (int i = 0; i < m; ++i) {
39                 for (int j = 0; j < n; ++j) {
40                     if (M[i][j] == 0)
41                         continue;
42                     for (int k = 0; k < 4; ++k) {
43                         int cnt = 0, x = i, y = j;
44                         while (x >= 0 && x < m && y >= 0 && y < n && M[x][y] == 1) {
45                             x += dirs[k][0];
46                             y += dirs[k][1];
47                             ++cnt;
48                         }
49                         res = Math.max(res, cnt);
50                     }
51                 }
52             }
53             return res;
54         }
55     }