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.

一刷
题解：我们可以用dfs求解这道题。注意，为了避免重复地dfs，可以用visited这个来cache子序列的longest increasing path. 用dp来降低时间复杂度。

class Solution {
    int max;
    int m, n;
    int[][] dirs = {{0,1}, {0,-1}, {1,0}, {-1,0}};
    public int longestIncreasingPath(int[][] matrix) {
        m = matrix.length;
        if(matrix == null || m==0) return 0;
        n = matrix[0].length;
        int[][] cache = new int[m][n];
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                int len = bfs(matrix, i, j, cache);
                max = Math.max(len, max);
            }
        }
        return max;
    }
    
    private int bfs(int[][] matrix, int i, int j, int[][] cache){
        if(cache[i][j]!=0) return cache[i][j];
        int max = 1;
        for(int[] dir : dirs){
            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;
            int len = bfs(matrix, x, y, cache);
            max = Math.max(max, 1+len);
        }
        cache[i][j] = max;
        return max;
    }
}