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.
一刷
题解：我们可以用dfs求解这道题。注意，为了避免重复地dfs，可以用visited这个来cache子序列的longest increasing path. 用dp来降低时间复杂度。
class Solution {
int max;
int m, n;
int[][] dirs = {{0,1}, {0,-1}, {1,0}, {-1,0}};
public int longestIncreasingPath(int[][] matrix) {
m = matrix.length;
if(matrix == null || m==0) return 0;
n = matrix[0].length;
int[][] cache = new int[m][n];
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
int len = bfs(matrix, i, j, cache);
max = Math.max(len, max);
}
}
return max;
}
private int bfs(int[][] matrix, int i, int j, 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 || y<0 || x>=m || y>=n || matrix[x][y]<=matrix[i][j]) continue;
int len = bfs(matrix, x, y, cache);
max = Math.max(max, 1+len);
}
cache[i][j] = max;
return max;
}
}
