[leetcode] Simplify Path

题目：

Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

click to show corner cases.

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".

代码：

class Solution {public:    /***     * 注意几个细节：     * 1)重复连续出现的'/'，只按1个处理，即跳过重复连续出现的'/'；     * 2)如果路径名是"."，则不处理；     * 3)如果路径名是".."，则需要弹栈，如果栈为空，则不做处理；     * 4)如果路径名为其他字符串，入栈。     *      *      * */    string simplifyPath(string path) {        int size = path.size();        stack<string> ss;        for(int i = 0; i < size;){            //跳过'/'            while('/' == path[i] && i < size)                ++ i;            //记录路径名称            string s = "";            while('/' != path[i] && i < size)                s += path[i ++];            //对当前路径进行处理            if(s == ".." && !ss.empty())                ss.pop();            else if(s != "" && s!= "." && s != "..")                ss.push(s);        }        string res = "";        if(ss.empty())            return "/";        else{            //注意输出顺序            while(!ss.empty()){                res = "/" + ss.top() + res;                ss.pop();            }        }        return res;                }};