• 【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()
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  • 原文地址:https://www.cnblogs.com/seyjs/p/12590740.html
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