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【leetcode】Simplify Path

时间:2014-12-29 22:39:53      阅读:184      评论:0      收藏:0      [点我收藏+]

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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".
 
用一个堆栈来模拟路径的行为,遇到"."不操作,遇到".."退栈,其他情况都压入堆栈。
程序中先把路径分割出来,然后再处理
 
  1 class Solution {
  2 
  3 public:
  5     string simplifyPath(string path) {
  9         vector<string> spl;
 11         spliceStr(path,spl);   
 15         vector<string> stk;
 17         for(int i=0;i<spl.size();i++)
 19         {
 21             if(spl[i]=="..")
 23             {
 25                 if(!stk.empty())
 27                 {
 29                   stk.pop_back();  
 31                 }
 33             }
 35             else if(spl[i]==".")
 37             {
 39                 continue;
 41             }
 43             else
 45             {
 47                 stk.push_back(spl[i]);
 49             }
 51         }
 55         string result="";
 57         for(int i=0;i<stk.size();i++)
 59         {
 61             result+="/"+stk[i];
 63         }
 67         if(result=="")
 69         {
 71             result="/";
 73         }
 79         return result;
 83     }
 84 
 85    
 86 
 87    
 88 
 89     void spliceStr(string &path,vector<string> &result)
 91     {
 93         int pre_index=0;
 95         string tmp;
 97         for(int i=0;i<path.size();i++)
 99         {
101             if(path[i]==/)
103             {
105                 tmp=path.substr(pre_index,i-pre_index);
107                 if(tmp!="")
109                 {
111                    result.push_back(tmp);
113                 }
115                 pre_index=i+1;
117             }
119         }
122 
123         if(pre_index<path.size())
125         {
127             tmp=path.substr(pre_index,path.size()-pre_index);
129             if(tmp!="")
131             {
133                 result.push_back(tmp);
135             }
137         }
139     }
141 };

 

【leetcode】Simplify Path

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原文地址:http://www.cnblogs.com/reachteam/p/4192434.html

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