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.
class Solution {
int directions[][] = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } };
public int longestIncreasingPath(int[][] matrix) {
if (matrix.length == 0)
return 0;
int dp[][] = new int[matrix.length][matrix[0].length];
int max = 0;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
if (dp[i][j] == 0) {
dp[i][j] = maxPath(matrix, dp, i, j);
}
max = Math.max(dp[i][j], max);
}
}
return max;
}
public int maxPath(int[][] matrix, int dp[][], int x, int y) {
if (dp[x][y] != 0)
return dp[x][y];
dp[x][y] = 1;// Base case
for (int i = 0; i < 4; i++) {
int newX = x + directions[i][0], newY = y + directions[i][1];
if (newX >= 0 && newX < matrix.length && newY >= 0 && newY < matrix[0].length
&& matrix[x][y] < matrix[newX][newY])
dp[x][y] = Math.max(dp[x][y], maxPath(matrix, dp, newX, newY) + 1);
}
return dp[x][y];
}
}