【LeetCode解答】Simplify Path

这题比较elegant的解法是弄个状态机什么的吧  遇到什么字符进什么状态触发什么操作之类的...  懒得弄那么复杂了直接if...else了...

class Solution {
public:
    string simplifyPath(string path) {
        string re;
        if(path.empty()) return re;
        re.push_back('/');
        for(int i=0;i<path.length();)
        {
            if(i<(int)path.length()-3)
            {
                if(0==strcmp(path.substr(i,3).c_str(),"/./"))
                    {if(re[re.length()-1]!='/')re.push_back('/');i+=2;continue;}
                if(0==strcmp(path.substr(i,4).c_str(),"/../"))
                    {if(re.length()>1)do{re.pop_back();}while(re[re.length()-1]!='/');i+=3;continue;}
            }
            else if(i==path.length()-3)
            {
                if(0==strcmp(path.substr(i,3).c_str(),"/./")) break;
                if(0==strcmp(path.substr(i,3).c_str(),"/..")) {if(re.length()>1)do{re.pop_back();}while(re[re.length()-1]!='/');break;}
            }
            else if(i==path.length()-2)
            {
                if(0==strcmp(path.substr(i,2).c_str(),"//")||0==strcmp(path.substr(i,2).c_str(),"/.")) break;
            }
            else if(i==path.length()-1)
            {
                if('/'==path[i]) break;
            }

            if(re[re.length()-1]!='/'||path[i]!='/')re.push_back(path[i]);
            i++;
        }
        if(re.length()>1&&'/'==re[re.length()-1]) re.pop_back();
        return re;
    }
};