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.


## Solutions

### 1. DFS+DP+记忆化-递归

使用 DFS 图搜索，注意记忆化！

# Time Complexity: O(mn)
# Space Complexity: O(mn)
class Solution:
def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
if not matrix:
return 0
r, c = len(matrix), len(matrix[0])
res = 1
dp = [[0 for _ in range(c)] for _ in range(r)]
for i in range(r):
for j in range(c):
dp[i][j] = self.dfs(matrix, i, j, r, c, dp)
res = max(res, dp[i][j])
return res

def dfs(self, matrix, i, j, r, c, dp):
if dp[i][j]:
return dp[i][j]
max_val = 1
for m, n in {(i+1, j), (i-1, j), (i, j+1), (i, j-1)}:
if m < 0 or m >= r or n < 0 or n >= c:
continue

if matrix[m][n] > matrix[i][j]:
# find the max val in the four directions
max_val = max(max_val, self.dfs(matrix, m, n, r, c, dp)+1)
dp[i][j] = max_val
return dp[i][j]
# Runtime: 512 ms, faster than 67.50% of Python3 online submissions for Longest Increasing Path in a Matrix.
# Memory Usage: 12.9 MB, less than 100.00% of Python3 online submissions for Longest Increasing Path in a Matrix.