3.14 Simplify Path

http://www.cnblogs.com/TenosDoIt/p/3465328.html

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".

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分析：需要注意的是/…可以表示名字为…的路径，路径的最后可能没有/。可以利用栈，碰到正常路径压入栈中，碰到/.不作任何操作，碰到/..删除栈顶元素。下面代码中用数组来模拟栈                                        

class Solution {
public:
    string simplifyPath(string path) {
        int len = path.size();
        vector<string> vec;
        int i = 0, index = 0;
        while(i < len)
        {
            int j = path.find('//', i + 1);
            string tmp;
            if(j != string::npos)
                tmp = path.substr(i, j - i);
            else {tmp = path.substr(i, len); j = len;}
         
            if(tmp == "/");
            else if(tmp == "/.");
            else if(tmp == "/..")
                {if(!vec.empty())vec.pop_back();}
            else
                vec.push_back(tmp);
            i = j;
        }
        if(vec.empty())return "/";
        else
        {
            string res;
            for(int i = 0; i < vec.size(); i++)
                res += vec[i];
            return res;
        }
    }
};