题目地址: https://leetcode.com/problems/longest-increasing-path-in-a-matrix/description/
题目描述：
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.
题目大意
求二维矩阵中最长的递增路径。
解题方法
和417. Pacific Atlantic Water Flowopen in new window非常类似，直接DFS求解。一般来说DFS需要有固定的起点，但是对于这个题，二维矩阵中的每个位置都算作起点。
把每个位置都当做起点，然后去做个dfs，看最长路径是多少。然后再找出全局的最长路径。使用cache保存已经访问过的位置，这样能节省了很多搜索的过程，然后有个continue是为了剪枝。因为这个做法比较暴力，就没有什么好讲的了。
最坏情况下的时间复杂度是O((MN)^2)，空间复杂度是O(MN)。
class Solution(object):
def longestIncreasingPath(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: int
"""
if not matrix or not matrix[0]:
return 0
m, n = len(matrix), len(matrix[0])
res = 0
cache = [[-1] * n for _ in range(m)]
for i in range(m):
for j in range(n):
path = self.dfs(matrix, cache, m, n, i, j)
res = max(res, path)
return res
def dfs(self, matrix, cache, m, n, i, j):
if cache[i][j] != -1:
return cache[i][j]
directions = [(-1, 0), (1, 0), (0, 1), (0, -1)]
res = 1
for dire in directions:
x, y = i + dire[0], j + dire[1]
if x < 0 or x >= m or y < 0 or y >= n or matrix[x][y] <= matrix[i][j]:
continue
path = 1 + self.dfs(matrix, cache, m, n, x, y)
res = max(path, res)
cache[i][j] = res
return cache[i][j]
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参考资料：
https://leetcode.com/problems/pacific-atlantic-water-flow/discuss/90739/Python-DFS-bests-85.-Tips-for-all-DFS-in-matrix-question./181815
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