LeetCode 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;
     }
 }