Given an absolute path for a file (Unix-style), simplify it.
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:http://blog.csdn.net/maverick1990/article/details/23275051
题意: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
C++实现代码:
#include<iostream>
#include<vector>
#include<string>
#include<sstream>
using namespace std;
class Solution
{
public:
string simplifyPath(string path)
{
if(path.empty())
return "";
vector<string> ret;
string tmp;
stringstream ss(path);
while(getline(ss,tmp,'/'))
{
if(tmp.empty()||tmp==".")
continue;
if(tmp=="..")
{
if(!ret.empty())
ret.pop_back();
}
else
ret.push_back(tmp);
}
tmp.clear();
for(int i=0; i<(int)ret.size(); i++)
{
tmp+="/";
tmp+=ret[i];
}
if(ret.empty())
return "/";
return tmp;
}
};
int main()
{
Solution s;
string ss="";
cout<<s.simplifyPath(ss)<<endl;
}
第二遍
class Solution { public: string simplifyPath(string path) { if(path.empty()) return ""; string res="/"; vector<string> vec; istringstream ss(path); string tmp; while(getline(ss,tmp,'/')) { if(tmp.empty()||tmp==".") continue; if(tmp==".."&&!vec.empty()) { vec.pop_back(); continue; } else if(tmp!="..") vec.push_back(tmp); } int i=0; while(i<vec.size()) { res+=vec[i]; ++i; if(i<vec.size()) res+='/'; } return res; } };