329. Longest Increasing Path in a Matrix

題目：
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.

解答：
最重要的是用cache保存每一個掃過結點的最大路徑。我開始作的時候，用全局變量記錄的max, dfs沒有返回值，這樣很容易出錯，由於任何一個用到max的環節都有可能改變max值，因此仍是在函數中定義，把當前的max直接返回計算不容易出錯。

public class Solution {
    public int DFS(int[][] matrix, int i, int j, int[][] cache) {
        if (cache[i][j] != 0) return cache[i][j];
        //改進：記錄下每個訪問過的點的最大長度，當從新訪問到這個點的時候，就不須要重複計算
        int[][] dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        int max = 1;
        for (int k = 0; k < dir.length; k++) {
            int x = i + dir[k][0], y = j + dir[k][1];
            if (x < 0 || x > matrix.length - 1 || y < 0 || y > matrix[0].length - 1) {
                continue;
            } else {
                if (matrix[x][y] > matrix[i][j]) {
                    //這裏當前結點只算長度1,而後加上dfs後子路徑的長度，比較得出最大值
                    int len = 1 + DFS(matrix, x, y, cache);
                    max = Math.max(max, len);
                }
            }
        }
        cache[i][j] = max;
        return max;
    }
    
    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 len = DFS(matrix, i, j, cache);
                max = Math.max(max, len);
            }
        }
        
        return max;
    }
}