题目：原题链接（困难）
标签：深度优先搜索、记忆化递归、拓扑排序
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N × M ) | O ( N × M ) | 552ms (42.32%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def __init__(self):
self.s1, self.s2 = 0, 0
self.matrix = [[]]
self.dp = [[]]
self.ans = 0
def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
if not matrix or not matrix[0]:
return 0
self.s1, self.s2 = len(matrix), len(matrix[0])
self.matrix = matrix
self.dp = [[-1] * self.s2 for _ in range(self.s1)]
for i1 in range(self.s1):
for i2 in range(self.s2):
self.dfs(i1, i2)
return self.ans
def dfs(self, i1, i2):
# 处理记忆化的情况
if self.dp[i1][i2] != -1:
return self.dp[i1][i2]
res = 0
for j1, j2 in self.get_next(i1, i2):
res = max(res, self.dfs(j1, j2))
res += 1
self.dp[i1][i2] = res
self.ans = max(self.ans, res)
return res
def is_valid(self, i1, i2):
return 0 <= i1 < self.s1 and 0 <= i2 < self.s2
def get_next(self, i1, i2):
res = []
for j1, j2 in [(i1 - 1, i2), (i1 + 1, i2), (i1, i2 - 1), (i1, i2 + 1)]:
if self.is_valid(j1, j2) and self.matrix[i1][i2] > self.matrix[j1][j2]:
res.append((j1, j2))
return res
