• 2019年3月6日 980. Unique Paths III


    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
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  • 原文地址:https://www.cnblogs.com/seenthewind/p/10483113.html
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