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"

.

思路：

利用堆栈来存储目录并完成消除。

代码：

class Solution {
public:
    stack<string> mystack;
    string simplifyPath(string path) {
        for(int i=0;i<path.size();i++)
        {
            //获得下一字符串
            string cur;
            if(path[i]=='/')
            {
                cur="/";
            }
            else
            {
                while(i<path.size()&&path[i]!='/')
                {
                    cur.push_back(path[i++]);
                }
                if(i<path.size())
                {
                    i--;
                }
            }
            //获得下一字符串
            
            if(mystack.size()==0)
            { 
                mystack.push(cur);
            }
            else
            {
                if(mystack.top()=="/"&&cur=="/")
                {
                    continue;
                }//忽略相邻的“//”
                else if(cur==".")
                {
                    continue; 
                }//忽略“.”
                else if(cur==".."&&mystack.top()=="/")
                {
                    mystack.pop();
                    while(mystack.size()>0&&mystack.top()!="/")
                    {
                        mystack.pop();
                    }
                }//返回上一层目录
                else
                {
                    mystack.push(cur);
                }
            }
        }
        string str;
        if(mystack.size()>1&&mystack.top()=="/")
        {
            mystack.pop();
        }//去掉最后一个“/”
        stack<string> reverse;
        while(mystack.size()>0)
        {
            reverse.push(mystack.top());
            mystack.pop();
        }//反转堆栈
        while(reverse.size()>0)
        {
            str+=reverse.top();
            reverse.pop();
        }//生成字符串
        if(str.size()==0)
            str="/";
        return str;
    }
};

另一种实现：

class Solution {
public:
    string simplifyPath(string path) {
        if(path.size() == 0){
            return "";
        }
        
        list<string> q;
        
        string cur;
        for(int i = 0; i < path.size() + 1; i++){
            if(i == path.size() || path[i] == '/'){
                if(cur.size() > 0){
                    if(cur == ".." && q.size() > 0 && q.back() != ".."){
                        q.pop_back();
                    }
                    else if(cur == ".." && q.size() == 0){
                    }
                    else if(cur != "."){
                        q.push_back(cur);
                    }
                    cur.clear();
                }
            }
            else{
                cur.push_back(path[i]);
            }
        }
        
        string result;
        for (std::list<string>::iterator it=q.begin(); it != q.end(); ++it){
            result.push_back('/');
            for(int j = 0; j < (*it).size(); j++){
                result.push_back((*it)[j]);
            }
        }
        
        if(result == ""){
            return "/";
        }
        return result;
    }
};