329. 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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

这道题给我们一个二维数组，让我们求矩阵中最长的递增路径，规定我们只能上下左右行走，不能走斜线或者是超过了边界。那么这道题的解法要用递归和DP来解，用DP的原因是为了提高效率，避免重复运算。我们需要维护一个二维动态数组dp，其中dp[i][j]表示数组中以(i,j)为起点的最长递增路径的长度，初始将dp数组都赋为0，当我们用递归调用时，遇到某个位置(x, y), 如果dp[x][y]不为0的话，我们直接返回dp[x][y]即可，不需要重复计算。我们需要以数组中每个位置都为起点调用递归来做，比较找出最大值。在以一个位置为起点用DFS搜索时，对其四个相邻位置进行判断，如果相邻位置的值大于上一个位置，则对相邻位置继续调用递归，并更新一个最大值，搜素完成后返回即可，参见代码如下：

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];
    }
}

　　矩阵常用的记忆化搜索, 返回值为基本数据类型

不要怀疑dfs的能力, 人家会一直递归再回溯给上一层正确的返回值