Longest Continuance Increasing sub-sequence in matrix

一个矩阵，求最长连续的序列长度

［1 2 3

4 5 6

7 8 9］

我的解法时用dfs遍历矩阵，如果便利过的元素就标记为false，不再遍历。

邻居就是上下左右

e.g.

1 2 3

6 5 4 -> 7

7 9 8

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

5 7 9

1 2 3 -> 3

4 6 8

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

9 8 7

4 5 6 ->9

3 2 1

题目一. 只能上下左右，走到边界处止步

题目二. 每一行／列都是一个循环，走到边界处譬如第 0 列，可以继续走到另一边界，譬如 n-1

 1 /*
 2  * Graph DFS
 3  * 
 4  * Find the Longest Continuance Increasing sub-sequence in 2d matrix
 5  * You could go up, down, left, right in the matrix
 6  * 
 7  */
 8 
 9 public class LIS2d {
10     public int findLongest(int[][] matrix) {
11         int max = 0;
12         int m = matrix.length, n = matrix[0].length;
13         boolean[][] visit = new boolean[m][n];
14         for (int i = 0; i < m; i++) {
15             for (int j = 0; j < n; j++) {
16                 int len = helper(matrix, visit, i, j, 1);
17                 max = max > len ? max : len;
18             }
19         }
20         return max;
21     }
22     public int helper(int[][] matrix, boolean[][] visit, int i, int j, int step) {
23         visit[i][j] = true;
24         int max = step;
25         int m = matrix.length, n = matrix[0].length;
26 //        if (i > 0 && matrix[i-1][j] == matrix[i][j]+1 && !visit[i-1][j]) {
27 //            int len = helper(matrix, visit, i-1, j, step+1);
28 //            max = max > len ? max : len;
29 //        }
30 //        if (i < m-1 && matrix[i+1][j] == matrix[i][j]+1 && !visit[i+1][j]) {
31 //            int len = helper(matrix, visit, i+1, j, step+1);
32 //            max = max > len ? max : len;
33 //        }
34 //        if (j > 0 && matrix[i][j-1] == matrix[i][j]+1 && !visit[i][j-1]) {
35 //            int len = helper(matrix, visit, i, j-1, step+1);
36 //            max = max > len ? max : len;
37 //        }
38 //        if (j < n-1 && matrix[i][j+1] == matrix[i][j]+1 && !visit[i][j+1]) {
39 //            int len = helper(matrix, visit, i, j+1, step+1);
40 //            max = max > len ? max : len;
41 //        }
42         int r = (i-1+m) % m;
43         int c = j;
44         if (matrix[r][c] == matrix[i][j]+1 && !visit[r][c]) {
45             int len = helper(matrix, visit, r, c, step+1);
46             max = max > len ? max : len;
47         }
48         r = (i+1+m) % m;
49         c = j;
50         if (matrix[r][c] == matrix[i][j]+1 && !visit[r][c]) {
51             int len = helper(matrix, visit, r, c, step+1);
52             max = max > len ? max : len;
53         }
54         r = i;
55         c = (j-1+n) % n;
56         if (matrix[r][c] == matrix[i][j]+1 && !visit[r][c]) {
57             int len = helper(matrix, visit, r, c, step+1);
58             max = max > len ? max : len;
59         }
60         r = i;
61         c = (j+1+n) % n;
62         if (matrix[r][c] == matrix[i][j]+1 && !visit[r][c]) {
63             int len = helper(matrix, visit, r, c, step+1);
64             max = max > len ? max : len;
65         }
66         visit[i][j] = false;
67         return max;
68     }
69     public static void main(String[] args) {
70         int[][] matrix = {{5, 6, 7},
71                           {4, 10, 9},
72                           {3, 2, 8}};
73         LIS2d lis = new LIS2d();
74         System.out.println(lis.findLongest(matrix));
75     }
76 }