Question

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.

Code

public class Solution {
    private int[] di = {1, 0, -1, 0};
    private int[] dj = {0, 1, 0, -1};
    private int max = 1;
    
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0) return 0;
        int m = matrix.length, n = matrix[0].length;
        int[][] dp = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                helper(dp, matrix, i, j);
            }
        }
        
        return max;
    }
    
    public int helper(int[][] dp, int[][] matrix, int i, int j) {
        if (dp[i][j] != 0) return dp[i][j];
        
        int m = matrix.length, n = matrix[0].length;
        
        dp[i][j] = 1;
        for (int k = 0; k < 4; k++) {
            int ni = i + di[k];
            int nj = j + dj[k];
            if (ni >= 0 && ni < m && nj >= 0 && nj < n) {
                if (matrix[i][j] > matrix[ni][nj]) {
                    dp[i][j] = Math.max(dp[i][j], helper(dp, matrix, ni, nj) + 1);
                }
            }
        }
        max = Math.max(max, dp[i][j]);
        
        return dp[i][j];
    }
}

Solution

DFS + 动态规划 + 记忆化搜索。

由于动态规划的初始起点不好确定，因此使用动态规划 + 记忆化搜索的方法完成。

dp[i][j]表示以matrix[i][j]为终点的最长递增路径的长度。通过DFS向四周搜索，如果matrix[四周i][四周j]小于matrix[i][j]，使用状态转移函数dp[i][j] = Math.max(dp[i][j], helper(dp, matrix, ni, nj) + 1); 最后返回dp中的最大值即可。