1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
| class Solution {
public int longestLine(int[][] mat) {
int ans = 0, m = mat.length, n = mat[0].length;
for (int i = 0; i < m; i++) {
int prev = 0;
for (int j = 0; j < n; j++) {
if (mat[i][j] == 1) {
prev += 1;
ans = Math.max(prev, ans);
} else {
prev = 0;
}
}
}
for (int i = 0; i < n; i++) {
int prev = 0;
for (int j = 0; j < m; j++) {
if (mat[j][i] == 1) {
prev += 1;
ans = Math.max(prev, ans);
} else {
prev = 0;
}
}
}
for (int i = 0; i < m; i++) {
int prev = 0, row = i, col = 0;
while (row < m && col < n) {
if (mat[row][col] == 1) {
prev += 1;
ans = Math.max(ans, prev);
} else {
prev = 0;
}
row += 1;
col += 1;
}
}
for (int i = 1; i < n; i++) {
int prev = 0, row = 0, col = i;
while (row < m && col < n) {
if (mat[row][col] == 1) {
prev += 1;
ans = Math.max(ans, prev);
} else {
prev = 0;
}
row += 1;
col += 1;
}
}
for (int i = 0; i < m; i++) {
int prev = 0, row = i, col = n - 1;
while (row < m && col >= 0) {
if (mat[row][col] == 1) {
prev += 1;
ans = Math.max(ans, prev);
} else {
prev = 0;
}
row += 1;
col -= 1;
}
}
for (int i = n - 2; i >= 0; i--) {
int prev = 0, row = 0, col = i;
while (row < m && col >= 0) {
if (mat[row][col] == 1) {
prev += 1;
ans = Math.max(ans, prev);
} else {
prev = 0;
}
row += 1;
col -= 1;
}
}
return ans;
}
}
|
Brute force. 先看horizontal, 在看vertical, 在看两种对角长度.
时间复杂度: O(4 * m * n) = O(mn)
空间复杂度: O(1)
