LintCode Longest Increasing Continuous subsequence II

原题链接在这里：http://www.lintcode.com/en/problem/longest-increasing-continuous-subsequence-ii/

题目：

Give you an integer matrix (with row size n, column size m)，find the longest increasing continuous subsequence in this matrix. (The definition of the longest increasing continuous subsequence here can start at any row or column and go up/down/right/left any direction).

Example

Given a matrix:

[
  [1 ,2 ,3 ,4 ,5],
  [16,17,24,23,6],
  [15,18,25,22,7],
  [14,19,20,21,8],
  [13,12,11,10,9]
]

return 25

题解：

DP. 状态是当前点为结束点，能有的最长延展长度. 转移方程是若周围四个方向有比我小的, 取符合条件的较大长度加1 dp[i][j] = Math.max(dp[dx][dy])+1.

但问题是如何按照由大到小的方向iterate 矩阵, 可以使用dfs.

Time Complexity: O(m * n), 每个点最多走两遍.

Space: O(m * n).

AC Java:

 1 public class Solution {
 2     public int longestIncreasingContinuousSubsequenceII(int[][] A) {
 3         if(A == null || A.length == 0|| A[0].length == 0){
 4             return 0;
 5         }
 6         
 7         int res = 1;
 8         int m = A.length;
 9         int n = A[0].length;
10         int [][] dp = new int[m][n];
11         boolean [][] visited = new boolean[m][n];
12         for(int i = 0; i<m; i++){
13             for(int j = 0; j<n; j++){
14                 dp[i][j] = dfs(A, i, j, visited, dp);
15                 res = Math.max(res, dp[i][j]);
16             }
17         }
18         return res;
19     }
20     
21     int [][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
22     private int dfs(int [][] A, int i, int j, boolean [][] visited, int [][] dp){
23         if(visited[i][j]){
24             return dp[i][j];
25         }
26         
27         int res = 1;
28         int m = A.length;
29         int n = A[0].length;
30         for(int [] dir : dirs){
31             int dx = i + dir[0];
32             int dy = j + dir[1];
33             if(dx>=0 && dx<m && dy>=0 && dy<n && A[i][j]>A[dx][dy]){
34                 res = Math.max(res, dfs(A, dx, dy, visited, dp)+1);
35             }
36         }
37         visited[i][j] = true;
38         dp[i][j] = res;
39         return res;
40     }
41 }