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".

Subscribe to see which companies asked this question.

解题技巧：

1. 重复连续出现的 '/'，只按1个处理，即：跳过重复连续出现的 '/'；

2. 如果路径名是 "."，则不处理；

3. 如果路径名是 ".."，则需要弹栈。如果栈为空，则不做处理；

4. 如果路径名为其他字符串，入栈。

5. 最终的结果是从栈中逐个取出元素，用'/'分隔并连接起来，不过要注意顺序。

代码：

#include <iostream>#include <stack>using namespace std;string simplifyPath(string path){     int len = path.size();     stack<string> stack1;     for(int i = 0; i < len; i++)     {         //跳过'/'         while(i < len && path[i] == '/')            i++;         //提取两个'/'之间的字符串         string tmp = "";         while(i < len && path[i] != '/')            tmp = tmp + path[i++];         //进行处理         if(tmp == ".")         {             continue;         }         else if(tmp == ".." )         {             if(!stack1.empty())                    stack1.pop();         }         else if(tmp != "")         {             stack1.push(tmp);         }     }     if(stack1.empty()) return "/";     string res = "";     while(!stack1.empty())     {         res = "/" + stack1.top()+res;         stack1.pop();     }     return res;}int main(){    string path;    cin>>path;    cout<<simplifyPath(path);}