LeetCode Day57 simplify path

按照’/\'将字符串分割，对每个分割出的字符串处理，如果是“…”，则popback，如果是“.”，则跳过，其他则保存。
最后输出结果
我的版本

class Solution {
public:
    string simplifyPath(string path) {
        vector<string> v;
        int flag=0,n=0;
        string s=\"\";
        for(int i=0;i<path.size();i++){
            if(path[i]==\'/\'){
                if(flag==0) continue;
                else {
                    flag=0;
                    if(s==\"..\"){
                        if(!v.empty()) v.pop_back();}
                    else if(s!=\".\") v.push_back(s);
                    s=\"\";
                    continue;
                }
            }
            else{
                s+=path[i];
                flag=1;
            }
        }
        if(s==\"..\"&&!v.empty()) v.pop_back();
        else if(s!=\"\"&&s!=\".\"&&s!=\"..\") v.push_back(s);
        if(v.empty()) return \"/\";
        string res;
        for(int i=0;i<v.size();i++){
            res+=(\'/\'+v[i]);
        }
        return res;
    }
};

class Solution {
public:
    string simplifyPath(string path) {
        vector<string> v;
        int i = 0;
        while (i < path.size()) {
            while (path[i] == \'/\' && i < path.size()) ++i;
            if (i == path.size()) break;
            int start = i;
            while (path[i] != \'/\' && i < path.size()) ++i;
            int end = i - 1;
            string s = path.substr(start, end - start + 1);
            if (s == \"..\") {
                if (!v.empty()) v.pop_back(); 
            } else if (s != \".\") {
                v.push_back(s);
            }
        }
        if (v.empty()) return \"/\";
        string res;
        for (int i = 0; i < v.size(); ++i) {
            res += \'/\' + v[i];
        }
        return res;
    }
};

简写版

class Solution {
public:
    string simplifyPath(string path) {
        string res, t;
        stringstream ss(path);
        vector<string> v;
        while (getline(ss, t, \'/\')) {
            if (t == \"\" || t == \".\") continue;
            if (t == \"..\"){
                if(!v.empty()) v.pop_back();}
            else v.push_back(t);
        }
        for (string s : v) res += \"/\" + s;
        return res.empty() ? \"/\" : res;
    }
};