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Simplify Path

Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

click to show corner cases.

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;}

 

Simplify Path