Longest Increasing Path in a Matrix

Difficulty: Medium
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

记忆化搜索

• DFS + 记忆化保存

public class Solution {

    public int longestIncreasingPath(int[][] matrix) {

        int m = matrix.length;
        if (m == 0) {
            return 0;
        }
        int n = matrix[0].length;
        if (n == 0) {
            return 0;
        }

        int[][] visited = new int[m][n];

        //Java会为数组自动初始化,所以初始化可以不用写
        // for (int i = 0; i < m; i++) {
        //     for (int j = 0; j < n; j++) {
        //         visited[i][j] = 0;
        //     }
        // }

        int max = 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                max = Math.max(max, dfs(i, j, m, n, matrix, visited));
            }
        }
        return max;
    }

    public int dfs(int i, int j, int m, int n, int[][] matrix, int[][] visited) {
        if (visited[i][j] != 0) {
            return visited[i][j];
        }

        int[] x = new int[]{0, 1, 0, -1};
        int[] y = new int[]{-1, 0, 1, 0};

        int max = 0;
        for (int k = 0; k < 4; k++) {
            int nx = i + x[k];
            int ny = j + y[k];
            if ( nx >= 0 && nx < m && ny >= 0 && ny < n && matrix[i][j] < matrix[nx][ny]) {
                max = Math.max(dfs(i + x[k], j + y[k], m, n, matrix, visited), max);
            }
        }
        visited[i][j] = max + 1;

        return visited[i][j];

    }
}