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

The longest increasing path is

`4`

The longest increasing path is

`[1, 2, 6, 9]`

.
Example 2:

nums = [ [3,4,5], [3,2,6], [2,2,1] ]

Return

The longest increasing path is

`4`

The longest increasing path is

`[3, 4, 5, 6]`

. Moving diagonally is not allowed.**Solution:**

This is a very classic DFS + memorialization problem. If we only use the DFS solution, it will end with many repeated calculations. Therefore, for each element in the matrix[i][j], we use a DP array dp[i][j] to denote the length of the maximum increasing path from this point. So along with the DFS, for a point in the matrix, if we've already found the longest increasing path, we don't have to repeatedly compute it again; we just need to return the length, which is dp[i][j].

One trick here is dp[i][j] stores the length of the longest increasing path. That is because the DFS from a point matrix[i][j] can guarantee the longest path from this point. Since we store this value in the dp[i][j], that can guarantee that dp[i][j] is the longest path from the point matrix[i][j].

**Code (Java):**

public class Solution { private int[] dx = new int[]{0, 0, -1, 1}; private int[] dy = new int[]{1, -1, 0, 0}; public int longestIncreasingPath(int[][] matrix) { if (matrix == null || matrix.length == 0) { return 0; } int m = matrix.length; int n = matrix[0].length; int max = 0; int[][] dp = new int[m][n]; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { max = Math.max(max, helper(i, j, matrix, dp)); } } return max; } private int helper(int row, int col, int[][] matrix, int[][] dp) { if (dp[row][col] > 0) { return dp[row][col]; } int m = matrix.length; int n = matrix[0].length; int curMax = 0; for (int i = 0; i < 4; i++) { int x = dx[i] + row; int y = dy[i] + col; if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[row][col]) { curMax = Math.max(curMax, helper(x, y, matrix, dp)); } } dp[row][col] = curMax + 1; return curMax + 1; } }

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