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".

Analyse: We need to understand the meaning of these symbols:

  "/" is the directory seperator. The content between two seperators have following meaning.

  "/." is the current directory, when it's encountered, we just ignore it.

  "/.." is the parent directory, if it's encountered and the stack is not empty, we pop the top element. 

  When other symbols are encountered, we just push into the stack. If the stack is empty, return "/".

 

After we coping with the string, we find that we want to pop the element from front to end. Thus, we use list structure. 

Runtime: 8ms.

复制代码
 1 class Solution {
 2 public:
 3     string simplifyPath(string path) {
 4         list<string> li;
 5         if(path.size() == 0) return "";
 6         
 7         int index = 0;
 8         while(index < path.size()){
 9             int comp = index; //judge whether there are elements between two seperators
10             string temp;
11             while(index < path.size() && path[index] != '/'){//extract elements between two seperators
12                 temp += path[index++];
13             }//index < path.size() is very important!! to avoid unlimited iteration
14             
15             if(index != comp){
16                 if(temp == ".."){
17                     if(!li.empty()) li.pop_back();
18                 }
19                 else if(temp == ".") continue;
20                 else li.push_back(temp);
21             }
22             index++;
23         }
24         string result;
25         while(!li.empty()){
26             result += "/" + li.front();
27             li.pop_front();
28         }
29         if(result.size() == 0) return "/";
30         return result;
31     }
32 };
复制代码

 

posted @   amazingzoe  阅读(138)  评论(0编辑  收藏  举报
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