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
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.
Note
memo -> result for Node(i, j)
each Node we apply the dfs search
Time complexity :O(mn). Each vertex/cell will be calculated once and only once, and each edge will be visited once and only once. The total time complexity is thenO(V+E).VVis the total number of vertices andEEis the total number of edges. In our problem,O(V)=O(mn),O(E) = O(4V) = O(mn).
Space complexity :O(mn). The cache dominates the space complexity.
Code
classSolution {publicintlongestIncreasingPath(int[][] matrix) {if (matrix ==null||matrix.length==0|| matrix[0] ==null|| matrix[0].length==0) {return0; }int res =0, m =matrix.length, n = matrix[0].length;int[][] memo =newint[m][n]; //res from Node (i,j)for (int i =0; i < m; i++) {for (int j =0; j < n; j++) { res =Math.max(res,dfs(matrix, i, j, memo)); } }return res; }privateintdfs(int[][] matrix,int i,int j,int[][] memo) {if (memo[i][j] !=0) {return memo[i][j]; }int m =matrix.length, n = matrix[0].length;int[][] dir =newint[][]{{1,0},{0,1},{-1,0},{0,-1}};for (int k =0; k <dir.length; k++) {int x = dir[k][0] + i;int y = dir[k][1] + j;if (x >=0&& x < m && y >=0&& y < n && matrix[x][y] > matrix[i][j]) { memo[i][j] =Math.max(memo[i][j],dfs(matrix, x, y, memo)); } } memo[i][j] +=1;return memo[i][j]; }}