[leetcode] Simplify Path
2014-09-09 13:06
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Given an absolute path for a file (Unix-style), simplify it.
For example,
path =
path =
Corner Cases:
Did you consider the case where path =
In this case, you should return
Another corner case is the path might contain multiple slashes
such as
In this case, you should ignore redundant slashes and return
参考链接:http://blog.csdn.net/maverick1990/article/details/23275051
思路:
简化一个Unix文件路径,注意:
(1)"/." 表示本级目录,可直接忽略
(2)"/.." 表示返回上一级目录,即若上一级目录存在,连同"/.."一并删除,否则只删除/..
(3)若去除冗余后路径为空,返回"/"
(4)若包含连续"/",删除多余的
(5)若路径不是单个“/”,删除路径最后一个“/”
分析:一开始用逐个字符判断的方法,考虑很多边界条件。后来用字符串分割的思想,比较简明。思路如下:
(1)用“/”分割字符串,遍历每个分割部分,存入一个vector<string>中
(2)若当前分割部分为空,证明有连续的"/"或是最后一个“/”,忽略
(3)若当前部分为“.”,忽略
(4)若当前部分为“..”,若vector不为空,去除vector最后一个元素
(5)再将vector中的string用“/”连起来,得到结果
代码:
For example,
path =
"/home/", =>
"/home"
path =
"/a/./b/../../c/", =>
"/c"
Corner Cases:
Did you consider the case where path =
"/../"?
In this case, you should return
"/".
Another corner case is the path might contain multiple slashes
'/'together,
such as
"/home//foo/".
In this case, you should ignore redundant slashes and return
"/home/foo".
参考链接:http://blog.csdn.net/maverick1990/article/details/23275051
思路:
简化一个Unix文件路径,注意:
(1)"/." 表示本级目录,可直接忽略
(2)"/.." 表示返回上一级目录,即若上一级目录存在,连同"/.."一并删除,否则只删除/..
(3)若去除冗余后路径为空,返回"/"
(4)若包含连续"/",删除多余的
(5)若路径不是单个“/”,删除路径最后一个“/”
分析:一开始用逐个字符判断的方法,考虑很多边界条件。后来用字符串分割的思想,比较简明。思路如下:
(1)用“/”分割字符串,遍历每个分割部分,存入一个vector<string>中
(2)若当前分割部分为空,证明有连续的"/"或是最后一个“/”,忽略
(3)若当前部分为“.”,忽略
(4)若当前部分为“..”,若vector不为空,去除vector最后一个元素
(5)再将vector中的string用“/”连起来,得到结果
代码:
class Solution { public: string simplifyPath(string path) { string ans,now; vector<string> list; stringstream ss(path); while(getline(ss,now,'/')){ if(now.size()==0 || now==".") continue; if(now==".."){ if(!list.empty()) list.pop_back(); } else{ list.push_back(now); } } for(int i=0;i<list.size();i++){ ans+="/"; ans+=list[i]; } if(ans.size()==0) ans="/"; return ans; } };
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