[LeetCode] 329. Longest Increasing Path in a Matrix

Problem

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:

Input: nums = 
[
  [9,9,4],
  [6,6,8],
  [2,1,1]
] 
Output: 4 

Explanation: The longest increasing path is [1, 2, 6, 9].
Example 2:

Input: nums = 
[
  [3,4,5],
  [3,2,6],
  [2,2,1]
] 
Output: 4 

Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

Solution

class Solution {
    private static final int[][] dirs = {{0,1},{0,-1},{-1,0},{1,0}};
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
        int m = matrix.length, n = matrix[0].length;
        int[][] cache = new int[m][n];
        int max = 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int curMax = dfs(matrix, i, j, cache);
                max = Math.max(max, curMax);
            }
        }
        return max;
    }
    
    private int dfs(int[][] matrix, int i, int j, int[][] cache) {
        //if saved(visited), return directly
        if (cache[i][j] != 0) return cache[i][j];
        int m = matrix.length, n = matrix[0].length;
        int max = 1;
        //this for loop is actually getting dfs result for 4 directions
        for (int[] dir: dirs) {
            int x = i+dir[0], y = j+dir[1];
            if (x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] <= matrix[i][j]) continue;
            int curMax = 1+dfs(matrix, x, y, cache);
            max = Math.max(max, curMax);
        }
        cache[i][j] = max;
        return max;
    }
}