Longest Increasing Path in a Matrix(Hard)
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).
给出一个数字矩阵，寻找一条最长上升路径，每个位置只能向上下左右四个位置移动。
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.
public static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
public int longestIncreasingPath(int[][] matrix) {
if(matrix.length == 0) return 0;
int rows = matrix.length, cols = matrix[0].length;
int[][] cache = new int[rows][cols];
int max = 1;
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++) {
int len = dfs(matrix, i, j, rows, cols, cache);
max = Math.max(max, len);
}
}
return max;
}
public int dfs(int[][] matrix, int i, int j, int rows, int cols, int[][] cache) {
if(cache[i][j] != 0) return cache[i][j];
int max = 1;
for(int[] dir: dirs) {
int x = i + dir[0];
int y = j + dir[1];
if(x < 0 || x >= rows || y < 0 || y >= cols || matrix[x][y] <= matrix[i][j])
continue;
int len = 1 + dfs(matrix, x, y, rows, cols, cache);
max = Math.max(max, len);
}
cache[i][j] = max;
return max;
}