LeetCode 329. Longest Increasing Path in a Matrix

原题链接在这里：https://leetcode.com/problems/longest-increasing-path-in-a-matrix/

题目：

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.

题解：

Longest increasing path could start from any grid of matrix.

We could perform DFS starting from each grid of matrix. For each of 4 neighbors, if it is within boundary and number > current grid number, continue DFS.

Could use memoization to save time. If this grid has already calcualted maximum increasing path before, simply return it.  

Time Complexity: O(mn). For DFS, it could take O(mn) time. But here use memoization, each grid could be visited no more than 5 times. m = matrix.length. n = matrix[0].length.

Space: O(m*n).用了memoization.

AC Java:

class Solution {
    HashMap<Integer, Integer> hm = new HashMap<>();
    int [][] dirs = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    
    public int longestIncreasingPath(int[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return 0;
        }
        
        int m = matrix.length;
        int n = matrix[0].length;
        int res = 0;
        for(int i = 0; i<m; i++){
            for(int j = 0; j<n; j++){
                res = Math.max(res, dfs(matrix, m, n, i, j));
            }
        }
        
        return res;
    }
    
    private int dfs(int [][] matrix, int m, int n, int i, int j){
        int key = i*n+j;
        if(hm.containsKey(key)){
            return hm.get(key);
        }
        
        int longest = 0;
        for(int [] dir : dirs){
            int x = i + dir[0];
            int y = j + dir[1];
            if(x>=0 && x<m && y>=0 && y<n && matrix[x][y]>matrix[i][j]){
                longest = Math.max(longest, dfs(matrix, m, n, x, y));   
            }
        }
        
        hm.put(key, longest+1);
        return longest+1;
    }
}