DFS+Cache

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:
Input: nums =
[
[9,9,4],
[6,6,8],
[2,1,1]
]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].
Example 2:
Input: nums =
[
[3,4,5],
[3,2,6],
[2,2,1]
]
Output: 4
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

public static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

public int longestIncreasingPath(int[][] matrix) {
    if(matrix.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, m, n, cache);
            max = Math.max(max, len);
        }
    }   
    return max;
}

public int dfs(int[][] matrix, int i, int j, int m, int n, int[][] cache) {
    if(cache[i][j] != 0) return cache[i][j];
    int max = 1;
    for(int[] dir: dirs) {
        int x = i + dir[0], y = j + dir[1];
        if(x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] <= matrix[i][j]) continue;
        int len = 1 + dfs(matrix, x, y, m, n, cache);
        max = Math.max(max, len);
    }
    cache[i][j] = max;
    return max;
}

注意：Cache的思想，The key is to cache the distance because it's highly possible to revisit a cell