LeetCode 980. Unique Paths III 使用位运算缓存路径

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最新推荐文章于 2024-07-14 02:25:25 发布

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该博客讨论了一个二维网格问题，其中包含起点、终点、空地和障碍。目标是计算从起点到终点的所有路径，条件是路径必须经过每个非障碍空地恰好一次。通过位运算解决这个问题，给出了具体的解决方案和示例。代码实现了一个名为`uniquePathsIII`的函数，用于计算符合条件的路径数量。

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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

-----------------

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]]))