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.

Solution：DFS + 记忆化搜索

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

class Solution {
    public static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

    public int longestIncreasingPath(int[][] matrix) {
        if(matrix.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 len = dfs(matrix, m, n, i, j, cache);
                max = Math.max(max, len);
            }
        }   
        return max;
    }

    public int dfs(int[][] matrix, int m, int n, int i, int j, int[][] cache) {
        if(cache[i][j] != 0) return cache[i][j];
        int max = 0;
        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 len = dfs(matrix, m, n, x, y, cache);
            max = Math.max(max, len);
        }
        cache[i][j] = max + 1;
        return max + 1;
    }
}