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.

我们采用DFS+memory的方法，就是在DFS的同时，记录当前元素所能构成的最大长度，如果下次再访问到这个点的时候直接返回这个点在memory中的值就可以了。时间复杂度为O(m*n),空间复杂度也是O(m*n)。代码如下：

public class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
        int[][] memory = new int[matrix.length][matrix[0].length];
        int max = 1;
        for(int i = 0; i < matrix.length; i++) {
            for(int j = 0; j < matrix[0].length; j++) {
                max = Math.max(max, getLength(i, j, Integer.MIN_VALUE, memory, matrix));
            }
        }
        return max;
    }
    
    public int getLength(int i, int j, int min, int[][] memory, int[][] matrix) {
        if(i < 0 || j < 0 || i == matrix.length || j == matrix[0].length || matrix[i][j] <= min)
            return 0;
        if(memory[i][j] != 0) 
            return memory[i][j];
        min = matrix[i][j];
        int a = getLength(i - 1, j, min, memory, matrix) + 1;
        int b = getLength(i + 1, j, min, memory, matrix) + 1;
        int c = getLength(i, j + 1, min, memory, matrix) + 1;
        int d = getLength(i, j - 1, min, memory, matrix) + 1;
        memory[i][j] = Math.max(a, Math.max(b, Math.max(c, d)));
        return memory[i][j]; 
    }
}