【leetcode】1391. Check if There is a Valid Path in a Grid

题目如下：

Given a m x n grid. Each cell of the grid represents a street. The street of grid[i][j] can be:

• 1 which means a street connecting the left cell and the right cell.
• 2 which means a street connecting the upper cell and the lower cell.
• 3 which means a street connecting the left cell and the lower cell.
• 4 which means a street connecting the right cell and the lower cell.
• 5 which means a street connecting the left cell and the upper cell.
• 6 which means a street connecting the right cell and the upper cell.

You will initially start at the street of the upper-left cell (0,0). A valid path in the grid is a path which starts from the upper left cell (0,0) and ends at the bottom-right cell (m - 1, n - 1). The path should only follow the streets.

Notice that you are not allowed to change any street.

Return true if there is a valid path in the grid or false otherwise.

Example 1:

Input: grid = [[2,4,3],[6,5,2]]
Output: true
Explanation: As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).

Example 2:

Input: grid = [[1,2,1],[1,2,1]]
Output: false
Explanation: As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)

Example 3:

Input: grid = [[1,1,2]]
Output: false
Explanation: You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).

Example 4:

Input: grid = [[1,1,1,1,1,1,3]]
Output: true

Example 5:

Input: grid = [[2],[2],[2],[2],[2],[2],[6]]
Output: true 

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• 1 <= grid[i][j] <= 6

解题思路：本题的关键在于确定移动方向，比如1向左移动可以对接上1，4，6；4向下移动可以对接上2，5，6。另外就是进出的方向，比如从5的左边进去，就只能从上面出去。如果存在符合题目要求的路径，那么这个路径一定是唯一的，确定了这些，只要从起点开始依次尝试即可。

代码如下：

class Solution(object):
    def hasValidPath(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: bool
        """
        dic_pair = {}
        dic_pair[(1,'L')] = [1,4,6]
        dic_pair[(1, 'R')] = [1,3,5]
        dic_pair[(2,'D')] = [2,5,6]
        dic_pair[(2,'U')] = [2,3,4]
        dic_pair[(3,'L')] = [1,4,6]
        dic_pair[(3,'D')] = [2,5,6]
        dic_pair[(4, 'R')] = [1,3,5]
        dic_pair[(4, 'D')] = [2,5,6]
        dic_pair[(5, 'U')] = [2,3,4]
        dic_pair[(5, 'L')] = [1,4, 6]
        dic_pair[(6, 'U')] = [2,3,4]
        dic_pair[(6, 'R')] = [1,3,5]

        dic_dir = {}
        dic_dir[1] = ['L','R']
        dic_dir[2] = ['U', 'D']
        dic_dir[3] = ['L', 'D']
        dic_dir[4] = ['R', 'D']
        dic_dir[5] = ['L', 'U']
        dic_dir[6] = ['R', 'U']

        queue = []

        def verify():
            #queue = []

            dic_visit = {}
            dic_visit[(0,0)] = 1

            direction = {}
            direction['R'] = (0,1)
            direction['L'] = (0, -1)
            direction['U'] = (-1,0)
            direction['D'] = (1,0)

            while len(queue) > 0:
                x,y,d = queue.pop(0)
                #print x,y
                if x == len(grid)-1 and y == len(grid[0]) - 1:
                    return True

                x1,y1 = direction[d]

                if x1 + x >= 0 and x1 + x < len(grid) and y + y1 >= 0 and y + y1 < len(grid[0]) \
                        and grid[x1 + x][y1 + y] in dic_pair[(grid[x][y], d)] and (x1 + x, y1 + y) not in dic_visit:
                    if d == 'L':
                        reversed_d = 'R'
                    elif d == 'R':
                        reversed_d = 'L'
                    elif d == 'U':
                        reversed_d = 'D'
                    else:
                        reversed_d = 'U'
                    inx = dic_dir[grid[x1 + x][y1 + y]].index(reversed_d)
                    if inx == 0:
                        d1 = dic_dir[grid[x1 + x][y1 + y]][1]
                    else:
                        d1 = dic_dir[grid[x1 + x][y1 + y]][0]
                    queue.append((x1 + x, y1 + y, d1))
                    dic_visit[(x1 + x, y1 + y)] = 1
            return False

        if grid[0][0] == 1:
            queue.append((0, 0, 'R'))
        elif grid[0][0] == 2:
            queue.append((0, 0, 'D'))
        elif grid[0][0] == 3:
            queue.append((0, 0, 'D'))
        elif grid[0][0] == 4:
            queue.append((0, 0, 'D'))
        elif grid[0][0] == 6:
            queue.append((0, 0, 'R'))
        return verify()