标签:位置 one 时间 描述 break border order iii ret
看描述就知道是一道搜索题,不过判断条件有点多,需要全部位置都踩一遍,相当于一笔画?
代码其实有点潦草了,因为额外的有点工作的事情,最近时间可能不多了。
class Solution(object): def checkPath(self, grid, n, m): for i in range(n): for j in range(m): if grid[i][j] == 0: return 0 return 1 def findPath(self, grid, i, j, n, m, path): mypath = copy.deepcopy(path) mygrid = copy.deepcopy(grid) mypath.append((i, j)) if grid[i][j] == 2: #print(mypath) return self.checkPath(grid, n, m) mygrid[i][j] = -1 #self.printGrid(grid, n, m) ret = 0 for vi, vj in ((0,1), (1,0), (-1,0), (0,-1)): x = i + vi y = j + vj if x >= 0 and y >= 0 and x < n and y < m: if mygrid[x][y] in (0, 2): ret += self.findPath(mygrid, x, y, n, m, mypath) return ret def uniquePathsIII(self, grid): ret = None n = len(grid) m = len(grid[0]) for i in range(n): for j in range(m): if grid[i][j] == 1: ret = self.findPath(grid, i, j, n, m, []) break if ret is not None: break return ret
2019年3月6日 980. Unique Paths III
标签:位置 one 时间 描述 break border order iii ret
原文地址:https://www.cnblogs.com/seenthewind/p/10483113.html