leetcode之Unique Paths II
2016-01-08 01:25
281 查看
这题相比于第一题就是增加了阻碍。不过我做I的时候的思路是以右下角为0,0,导致多做了好几遍无用功。反而速度不快了。可以改进下的其实。代码如下:
class Solution(object): def uniquePathsWithObstacles(self, obstacleGrid): """ :type obstacleGrid: List[List[int]] :rtype: int """ if obstacleGrid[len(obstacleGrid) - 1][len(obstacleGrid[0]) - 1] == 1: return 0 for i in range(len(obstacleGrid)): for j in range(len(obstacleGrid[0])): if obstacleGrid[i][j] == 0: obstacleGrid[i][j] = 1 else: obstacleGrid[i][j] = 0 #去除边界的阻碍的情况 for i in range(len(obstacleGrid))[::-1]: if obstacleGrid[i][-1] == 0: for j in range(i): obstacleGrid[j][-1] = 0 break for i in range(len(obstacleGrid[0]))[::-1]: if obstacleGrid[-1][i] == 0: for j in range(i): print j, i obstacleGrid[-1][j] = 0 break #去除中间有阻碍的情况 for i in range(len(obstacleGrid) - 1)[::-1]: for j in range(len(obstacleGrid[0]) - 1)[::-1]: if obstacleGrid[i][j] != 0: if obstacleGrid[i + 1][j] != 0 and obstacleGrid[i][j + 1] != 0: obstacleGrid[i][j] = obstacleGrid[i][j + 1] + obstacleGrid[i + 1][j] elif obstacleGrid[i + 1][j] == 0 and obstacleGrid[i][j + 1] == 0: obstacleGrid[i][j] = 0 elif obstacleGrid[i + 1][j] == 0: obstacleGrid[i][j] = obstacleGrid[i][j + 1] else: obstacleGrid[i][j] = obstacleGrid[i + 1][j] return obstacleGrid[0][0]
相关文章推荐
- Python动态类型的学习---引用的理解
- Python3写爬虫(四)多线程实现数据爬取
- 垃圾邮件过滤器 python简单实现
- 下载并遍历 names.txt 文件,输出长度最长的回文人名。
- install and upgrade scrapy
- Scrapy的架构介绍
- Centos6 编译安装Python
- 使用Python生成Excel格式的图片
- 让Python文件也可以当bat文件运行
- [Python]推算数独
- Python中zip()函数用法举例
- Python中map()函数浅析
- Python将excel导入到mysql中
- Python在CAM软件Genesis2000中的应用
- 使用Shiboken为C++和Qt库创建Python绑定
- FREEBASIC 编译可被python调用的dll函数示例
- Python 七步捉虫法