LeetCode OJ-- Simplify Path **

https://oj.leetcode.com/problems/simplify-path/

对linux路径的规范化，属于字符串处理的题目。细节很多。

#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
    string simplifyPath(string path) {
        if(path == "/../")
            return "/";

        int len = path.length();
        // if last was ..,then regard it as ../
        if(len>= && path[len-]=='.'&& path[len-] == '.')
            path.append("/");
        len = path.length();

        string ans = "/";
        int pos = ;
        int pos2;
        while()
        {
            //unfind is -1
            pos2 = path.find_first_of('/',pos);

            //two '//', ignore
            if(pos2!= - && path[pos2-]=='/')
            {
                pos = pos2+;
                continue;
            }
            //exit test,means not find
            //handle the left eg:/a/b/cc
            if(pos2 == - )
            {
                string subStr = path.substr(pos,pos2-pos+);
                if(subStr == "." || subStr == "..")
                    break;

                ans.append(subStr);
                break;
            }
            //if ../
            if(pos2>= && path[pos2-] == '.' && path[pos2-] == '.' && pos2- == pos)
            {
                int pos3 = ans.rfind('/',ans.size()-);
                if(pos3>=)
                    ans = ans.substr(,pos3+);
                pos = pos2 +;
                continue;
            }
            //if ./
            if(pos2>=&&path[pos2-] == '.'&& pos == pos2-)
            {
                pos = pos2+;
                continue;
            }

            string subStr = path.substr(pos,pos2 - pos +);
            ans.append(subStr);
            pos = pos2 + ;

        }
        //remove the last /
        int t = ans.length()-;
        while(t>)
        {
            if(ans[t]=='/')
                t--;
            else
                break;
        }

        ans = ans.substr(,t+);

        return ans;
    }
};

int main()
{
    class Solution myS;
    cout<<myS.simplifyPath("/b/DfZ/AT/ya///./../.././..")<<endl;
    return ;
}