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

 

/*
 * Graph DFS
 * 
 * Find the Longest Continuance Increasing sub-sequence in 2d matrix
 * You could go up, down, left, right in the matrix
 * 
 */

public class LIS2d {
    public int findLongest(int[][] matrix) {
        int max = 0;
        int m = matrix.length, n = matrix[0].length;
        boolean[][] visit = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int len = helper(matrix, visit, i, j, 1);
                max = max > len ? max : len;
            }
        }
        return max;
    }
    public int helper(int[][] matrix, boolean[][] visit, int i, int j, int step) {
        visit[i][j] = true;
        int max = step;
        int m = matrix.length, n = matrix[0].length;
//        if (i > 0 && matrix[i-1][j] == matrix[i][j]+1 && !visit[i-1][j]) {
//            int len = helper(matrix, visit, i-1, j, step+1);
//            max = max > len ? max : len;
//        }
//        if (i < m-1 && matrix[i+1][j] == matrix[i][j]+1 && !visit[i+1][j]) {
//            int len = helper(matrix, visit, i+1, j, step+1);
//            max = max > len ? max : len;
//        }
//        if (j > 0 && matrix[i][j-1] == matrix[i][j]+1 && !visit[i][j-1]) {
//            int len = helper(matrix, visit, i, j-1, step+1);
//            max = max > len ? max : len;
//        }
//        if (j < n-1 && matrix[i][j+1] == matrix[i][j]+1 && !visit[i][j+1]) {
//            int len = helper(matrix, visit, i, j+1, step+1);
//            max = max > len ? max : len;
//        }
        int r = (i-1+m) % m;
        int c = j;
        if (matrix[r][c] == matrix[i][j]+1 && !visit[r][c]) {
            int len = helper(matrix, visit, r, c, step+1);
            max = max > len ? max : len;
        }
        r = (i+1+m) % m;
        c = j;
        if (matrix[r][c] == matrix[i][j]+1 && !visit[r][c]) {
            int len = helper(matrix, visit, r, c, step+1);
            max = max > len ? max : len;
        }
        r = i;
        c = (j-1+n) % n;
        if (matrix[r][c] == matrix[i][j]+1 && !visit[r][c]) {
            int len = helper(matrix, visit, r, c, step+1);
            max = max > len ? max : len;
        }
        r = i;
        c = (j+1+n) % n;
        if (matrix[r][c] == matrix[i][j]+1 && !visit[r][c]) {
            int len = helper(matrix, visit, r, c, step+1);
            max = max > len ? max : len;
        }
        visit[i][j] = false;
        return max;
    }
    public static void main(String[] args) {
        int[][] matrix = {{5, 6, 7},
                          {4, 10, 9},
                          {3, 2, 8}};
        LIS2d lis = new LIS2d();
        System.out.println(lis.findLongest(matrix));
    }
}