Leetcode 71. Simplify Path

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".

简化Unix文件绝对路径，就是添加了许多.和/的无关字符。需要依次分析路径，遇到../向后回溯，注意判断以下几种特别情况。如果输入如以下序列：

"////b"
"/home/"
"/a/./b/.."
"/a/./b/../../c/../../c/a/d"
"/.....hidden"
"/..hidden"
"/a/b/d../.././.../..../...hidd..en.."
"/."
"./.."
"/../..."
".././a.."
"..."

那么输出为：

"/b"
"/home"
"/a"
"/c/a/d"
"/.....hidden"
"/..hidden"
"/a/b/.../..../...hidd..en.."
"/"
"/"
"/..."
"/a.."
"/"

C 代码已通过，如下所示：

char* simplifyPath(char* path) {
    int length=0,anslength=0;
    while(path[length]!='\0')
    {
        if(path[length]=='/')
        {
            path[anslength]=path[length];
            anslength++;
            break;
        }
        length++;
    }
    if(anslength==0)
    {
        path[0]='/';
        path[1]='\0';
        return path;
    }
    while(path[length]!='\0')
    {
        if(path[length]=='/')
        {
            if(path[anslength-1]!='/')
            {
                path[anslength]=path[length];
                anslength++;
                length++;
            }
            else
            {
                length++;
            }
        }
        else if(path[length]=='.')
        {
            if(path[length+1]=='\0'||(path[length+1]!='\0'&&path[length+1]=='/'))
            {
                length++;
            }
            else if(path[length+1]!='\0'&&path[length+1]=='.'&&(path[length+2]=='\0'||(path[length+2]!='\0'&&path[length+2]=='/')))
            {
                if(anslength>1)anslength--;
                while(anslength>1&&path[anslength-1]!='/')
                    anslength--;
                length=length+2;
            }
            else
            {
                while(path[length]!='\0'&&path[length]!='/')
                {
                    path[anslength]=path[length];
                    anslength++;
                    length++;
                }
            }
        }
        else
        {
            while(path[length]!='\0'&&path[length]!='/')
            {
                path[anslength]=path[length];
                anslength++;
                length++;
            }
        }
        //printf("%d %s\n",anslength,path);
    }
    //while(path[anslength-1]=='.')
   //     anslength--;
    if(anslength>1&&path[anslength-1]=='/')
        anslength--;
    path[anslength]='\0';
    return path;
}