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.
分析
memorize 用dp数组记录已经算过的,dp这里也起了visited的作用
几个边界条件
1visited dfs开始就用DP返回
2 x,y过界和没有递增,在for loop里continue掉
DFS
class Solution:
d = [-1,0,1,0,-1]
n=m=0
def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
if not matrix or not matrix[0]:
return 0
self.n,self.m=len(matrix),len(matrix[0])
res = 0
dp = [[0]*self.m for _ in range(self.n)]
for i in range(self.n):
for j in range(self.m):
res = max(res, self.dfs(matrix,dp,i,j))
return res
def dfs(self,matrix: List[List[int]],dp: List[List[int]], x:int, y:int) -> int:
if dp[x][y]:
return dp[x][y]
nxy = [(x+self.d[i],y+self.d[i+1]) for i in range(4) ]
maxStep = max([self.dfs(matrix,dp,nx,ny) for nx,ny in nxy if nx < self.n and nx>= 0 and ny < self.m and ny>=0 and matrix[nx][ny] > matrix[x][y] ] or [0])
dp[x][y] = 1 + maxStep
return 1+maxStep
class Solution:
d = [-1,0,1,0,-1]
n=m=0
def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
if not matrix or not matrix[0]:
return 0
self.n,self.m = len(matrix),len(matrix[0])
res = 0
cnt = self.m * self.n
while cnt:
visted = set()
for i in range(self.n):
for j in range(self.m):
if matrix[i][j] == float('-inf'):
continue
if (i == 0 or matrix[i][j] >= matrix[i-1][j]) and (i == self.n-1 or matrix[i][j] >= matrix[i+1][j]) and (j == 0 or matrix[i][j] >= matrix[i][j - 1]) and (j == self.m - 1 or matrix[i][j] >= matrix[i][j + 1]):
visted.add((i,j))
for i,j in visted:
matrix[i][j] = float('-inf')
cnt -= 1
res +=1
return res