# 329 Longest Increasing Path in a Matrix

### Problem:

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.

# Solutions:

``````public class Solution {
public int longestIncreasingPath(int[][] matrix) {
if( matrix == null || matrix.length==0 || matrix[0].length == 0) {
return 0;
}
int[] dx = {0, 1, 0, -1};
int[] dy = {1, 0, -1, 0};
int[][] maxs = new int[matrix.length][matrix[0].length];
int max = 0;
for(int i = 0; i < matrix.length; i++){
for(int j = 0; j < matrix[0].length; j++){
max = Math.max(max, dfs(dx, dy, matrix, i, j, maxs));
}
}
return max;
}

public int dfs(int[] dx, int[] dy, int[][] matrix, int i, int j,  int[][] maxs){
if(maxs[i][j] != 0) {
return maxs[i][j];
}
for(int m = 0; m < 4; m++){
int x = i + dx[m];
int y = j + dy[m];

if(x >= 0 && y >= 0 && x < matrix.length && y < matrix[0].length && matrix[x][y]>matrix[i][j]){
maxs[i][j] = Math.max(maxs[i][j], dfs(dx, dy, matrix, x, y, maxs));
}
}
maxs[i][j] ++;
return maxs[i][j];
}
}
``````