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".

一刷
题解：
先Split整个path by "/"， 然后建立一个stack / deque或者doubly linked list。 遍历split好的数组， 当数组当前元素p不为 ""， "."和".."的时候，我们push p 入栈， 否则当p为".."并且栈非空的情况下，我们pop上一个元素出栈。11ms

public class Solution {
   public String simplifyPath(String path) {
        if (path == null || path.length() < 2) return path;
        Stack<String> stack = new Stack<>();
        
        for (String p : path.split("/")) {
            if (!p.equals(".") && !p.equals("..") && !p.equals("")) stack.push(p);
            else if (p.equals("..") && !stack.isEmpty()) stack.pop(); 
        }
        
        StringBuilder res = new StringBuilder();
        while (!stack.isEmpty()) res.insert(0, stack.pop()).insert(0, "/");
        return res.length() == 0 ? "/" : res.toString();
    }
}

问题在于，对于StringBuilder,每次在开头insert时间复杂度很高。想办法从头部开始，可以考虑双向链表。10ms。感觉并没有很大提升。

public class Solution {
    public String simplifyPath(String path) {
        if (path == null || path.length() < 2) return path;
        LinkedList<String> list = new LinkedList<String>();
        
        for (String p : path.split("/")) {
            if (!p.equals(".") && !p.equals("..") && !p.equals("")) list.add(p);
            else if (p.equals("..") && !list.isEmpty()) list.remove(list.size()-1);
        }
        
        StringBuilder res = new StringBuilder();
        res.append("/");
        while (!list.isEmpty()) res.append(list.poll()).append("/");
        if(res.length()!=1) res.deleteCharAt(res.length()-1);
        return res.toString();
    }
}