Leetcode: Longest Increasing Path in a Matrix

Given an integer matrix, find the length of the longest increasing path.

From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).

Example 1:

nums = [
  [9,9,4],
  [6,6,8],
  [2,1,1]
]
Return 4
The longest increasing path is [1, 2, 6, 9].

Example 2:

nums = [
  [3,4,5],
  [3,2,6],
  [2,2,1]
]
Return 4
The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

DFS + DP:

use a two dimensional matrix dp[i][j] to store the length of the longest increasing path starting at matrix[i][j]

transferring function is: dp[i][j] = max(dp[i][j], dp[x][y] + 1), where dp[x][y] is its neighbor with matrix[x][y] > matrix[i][j]

Note:

1. Use matrix[x][y] > matrix[i][j] so we don't need a visited[m][n] array
2. The key is to cache the distance because it's highly possible to revisit a cell

Follow Up: How to get the actual longest increasing path

我的想法：类似Largest Divisible Subset, 除了一个dp[i][j]记录longest length以外，另外再用一个matrix pre[i][j]记录(i,j)longest increasing path上一跳位置, 并用一个variable记录最后最长的path的起始位置（第19行每次res更新时更新）。最后通过这个起始位置沿着一个一个上一跳位置，可以求出path

public class Solution {
    int[][] dp;
    int[][] directions = new int[][]{{-1,0},{1,0},{0,-1},{0,1}};
    int m;
    int n;

    public int longestIncreasingPath(int[][] matrix) {
        if (matrix==null || matrix.length==0 || matrix[0].length==0) return 0;
        m = matrix.length;
        n = matrix[0].length;
        dp = new int[m][n];

        int result = 0;

        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if (dp[i][j] == 0) 
                    dp[i][j] = DFS(i, j, matrix); 
                result = Math.max(result, dp[i][j]);
            }
        }
        return result;
    }
    
    public int DFS(int i, int j, int[][] matrix) {
        if (dp[i][j] != 0) return dp[i][j];
        dp[i][j] = 1;
        for (int[] dir : directions) {
            int x = i + dir[0];
            int y = j + dir[1];
            if (x<0 || y<0 || x>=m || y>=n || matrix[x][y]<=matrix[i][j]) continue;
            dp[i][j] = Math.max(dp[i][j], DFS(x, y, matrix)+1);
        }
        return dp[i][j];
    }
}