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描述
You are given an m x n integer array grid where grid[i][j] could be:
- 1 representing the starting square. There is exactly one starting square.
- 2 representing the ending square. There is exactly one ending square.
- 0 representing empty squares we can walk over.
- -1 representing 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: grid = [[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: grid = [[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: grid = [[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:
m == grid.length
n == grid[i].length
1 <= m, n <= 20
1 <= m * n <= 20
-1 <= grid[i][j] <= 2
There is exactly one starting cell and one ending cell.
解析
根据题意，就是给出了一个 MxN 大小的矩阵，每个格子可能有有四种数字表示不同的含义：
- 1 表示的是开始位置
- 2 表示的是结束为止
- 0 表示的是可以正常行走的空位置
- -1 表示的是无法跨越的障碍位置
题目要求我们要从开始位置走，到结束位置停下来，有多少种不同的走法可以将所有的空位置都经过一次。
有经验的通知一看这个找路径的题就知道得动态规划，但是这个题的 M 和 N 比较小，用回溯的思想也能找到可能路径，通过写递归函数来进行求解：
- 先对 grid 进行遍历，找到开始位置记为 [x,y] ，记录空格子的个数为 empty
- 使用递归函数 dfs 来搜索路径，这里面写法比较常规，就是 x 和 y 要在合法范围内进行上下左右的一步的走动，就是有一点技巧，为了递归时不重走来时的路，会暂时将当前的位置设置为 -1 ，在递归结束再恢复为 0 ，当走到结束位置且 empty== -1 表示空格都经过一次的时候才返回 1 ，否则其他情况都返回 0
- 递归结束得到结果就为答案
解答
class Solution(object):
def uniquePathsIII(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
def dfs(x,y,grid,empty):
if x<0 or x>=M or y<0 or y>=N:
return 0
if grid[x][y] == -1:
return 0
if grid[x][y] == 2:
if empty == -1:
return 1
return 0
grid[x][y] = -1
count = dfs(x-1, y, grid, empty-1) + dfs(x+1, y, grid, empty-1) + dfs(x, y-1, grid, empty-1) + dfs(x, y+1, grid, empty-1)
grid[x][y] = 0
return count
M = len(grid)
N = len(grid[0])
empty = 0
for i in range(M):
for j in range(N):
if grid[i][j] == 1:
x,y = i,j
elif grid[i][j] == 0:
empty += 1
return dfs(x,y,grid,empty)
运行结果
Runtime: 85 ms, faster than 14.29% of Python online submissions for Unique Paths III.
Memory Usage: 13.3 MB, less than 69.05% of Python online submissions for Unique Paths III.
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原题链接：leetcode.com/problems/un…
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