DFS经典题，reachable or not in a 2D maze
[[0, 0, 0, 0, 0, 1],
[1, 1, 0, 0, 0, 1],
[0, 0, 0, 1, 0, 0],
[0, 1, 1, 0, 0, 1],
[0, 1, 0, 0, 1, 0],
[0, 1, 0, 0, 0, 2]]
给定类似上面个一个 m by n matrix。其中0代表可以走的路，1代表不能走的墙。任意给出起点坐标(i, j)判段是否能从起点走到用2标出来的目标点？
1 grid = [[0, 0, 0, 0, 0, 1], 2 [1, 1, 0, 0, 0, 1], 3 [0, 0, 0, 1, 0, 0], 4 [0, 1, 1, 0, 0, 1], 5 [0, 1, 0, 0, 1, 0], 6 [0, 1, 0, 0, 0, 2]] 7 8 9 def dfsSearch(x, y): 10 11 if grid[x][y] == 2: 12 print('found at %d,%d' %(x, y)) 13 return True 14 elif grid[x][y] == 1: 15 print('Wall at %d,%d' %(x, y)) 16 elif grid[x][y] == 3: 17 print('Visited before: %d, %d' %(x, y)) 18 return False 19 20 print('Visiting %d,%d' %(x, y)) 21 22 # mark as visited 23 grid[x][y] = 3 24 25 # explore neighbors clockwise starting by the one on the right 26 if ((x < len(grid) - 1) and dfsSearch(x + 1, y)) or \ 27 (y > 0 and dfsSearch(x, y - 1)) or \ 28 (x > 0 and dfsSearch(x - 1, y)) or \ 29 ((y < len(grid) - 1) and dfsSearch(x, y + 1)): 30 return True 31 32 return False
这个算法只是检查能不能找到符合要求的path，但是不保证找到的那一条path是最短的。注意把已经访问过的点设置成3，这样访问过的点就不会被重复访问了。
如果想要找出shortest path的长度，可以用如下的方法：
1 class findShortestPath(object): 2 def __init__(self): 3 self.minLen = [0x7FFFFFFF] 4 5 def shortPath(self, x, y): 6 minLen = [999999] 7 self.shortPathHelper(x, y, 0) 8 return min(minLen) 9 10 def shortPathHelper(self, x, y, step): 11 nextStep = [[0, 1], [1, 0], [0, -1], [-1, 0]] 12 if grid[x][y] == 2: 13 print('found at %d,%d' % (x, y)) 14 self.minLen.append(step) 15 return True 16 elif grid[x][y] == 1: 17 print('Wall at %d,%d' % (x, y)) 18 elif grid[x][y] == 3: 19 print('Visited before: %d, %d' % (x, y)) 20 return False 21 22 23 grid[x][y] = 3 24 25 for k in range(4): 26 tx = x + nextStep[k][0] 27 ty = y + nextStep[k][1] 28 29 if (tx < 0 or tx > len(grid)-1 or ty < 0 or ty > len(grid)-1): 30 continue 31 self.shortPathHelper(tx, ty, step + 1)
