【leetcode】980. Unique Paths III

题目如下：

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. 1 <= grid.length * grid[0].length <= 20

解题思路：因为grid数据非常少，所以直接DFS/BFS即可得到答案。遍历grid的过程中记录每个节点是否已经遍历过，通过记录已经遍历了遍历节点的总数

代码如下：

class Solution(object):
    def uniquePathsIII(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        import copy
        visit = []
        count = 0
        total = len(grid) * len(grid[0])
        startx,starty = 0,0
        for i in range(len(grid)):
            visit.append([0] * len(grid[i]))
            for j in range(len(grid[i])):
                if grid[i][j] == -1:
                    count += 1
                elif grid[i][j] == 1:
                    startx,starty = i,j
        visit[startx][starty] = 1
        queue = [(startx,starty,copy.deepcopy(visit),1)]
        res = 0
        while len(queue) > 0:
            x,y,v,c = queue.pop(0)
            if grid[x][y] == 2 and c == total - count:
                res += 1
                continue
            direction = [(-1,0),(1,0),(0,1),(0,-1)]
            for i,j in direction:
                if x + i >= 0 and x + i < len(grid) and y + j >= 0 and y + j < len(grid[0]) and v[x+i][y+j] == 0 and grid[x+i][y+j] != -1:
                    v_c = copy.deepcopy(v)
                    v_c[x+i][y+j] = 1
                    queue.append((x+i,y+j,v_c,c+1))
        return res

标签：Paths,startx,square,980,len,ending,grid,Unique,over
来源： https://www.cnblogs.com/seyjs/p/10346330.html