On a 2-dimensional grid
, there are 4 types of squares:
1
represents the starting square. There is exactly one starting square.2
represents the ending square. There is exactly one ending square.0
represents empty squares we can walk over.-1
represents obstacles that we cannot walk over.
Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.
Example 1:
Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
Example 2:
Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)
Example 3:
Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.
Note:
1 <= grid.length * grid[0].length <= 20
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路径实际上是集合的一个问题,类似24点的场景,可以考虑用位运算,正好数据范围也够,上代码:
from typing import List
class Solution:
def uniquePathsIII(self, grid: List[List[int]]) -> int:
def pos_to_int(x, y, cols):
return 1 << (x * cols + y)
rows, cols = len(grid), len(grid[0]) if grid else 0
if rows == 0 or cols == 0:
return 0
target, sx, sy = 0, 0, 0
for i in range(rows):
for j in range(cols):
if grid[i][j] == 1:
sx, sy = i, j
if (grid[i][j] != -1):
target = target | pos_to_int(i, j, cols)
c, n, ct, res = 0, 1, pos_to_int(sx,sy,cols), 0
layers = [[(sx, sy, ct)],[]]
while (layers[c]):
for cx, cy, ct in layers[c]:
for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
nx, ny = cx + dx, cy + dy
if (0 <= nx < rows and 0 <= ny < cols and grid[nx][ny] != -1):
pos_int = pos_to_int(nx, ny, cols)
if (pos_int & ct == 0 and grid[nx][ny] == 2 and pos_int | ct == target):
res += 1
elif (pos_int & ct == 0 and grid[nx][ny] == 0):
nt = pos_int | ct
layers[n].append((nx, ny, nt))
layers[c].clear()
c, n = n, c
return res
s = Solution()
print(s.uniquePathsIII([[0,1],[2,0]]))