71. 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"

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".

链接：https://leetcode.com/problems/simplify-path/#/description

5/27/2017

9ms. 86%

注意的问题：

1. LinkedList当作stack的push, pop：操作在链表头。list.push(1), list.push(2)画图表示出来为2->1，所以此时list.get(0)返回的应该是2。不要想当然认为是从尾巴开始加，毕竟对于链表最方便的操作就是在表头。

2. 如果记不住前一点，就用removeLast, addLast来表示好了

3. 26行注意判断是否返回的是root

 1 public class Solution {
 2     public String simplifyPath(String path) {
 3         if (path == null || path.length() == 0 || path.equals("")) {
 4             return "";
 5         }
 6         LinkedList<String> list = new LinkedList<String>();
 7         String[] parts = path.split("/");
 8 
 9         for (int i = 0; i < parts.length; i++) {
10             if (parts[i].equals("") || parts[i].equals(".")) continue;
11             else if (parts[i].equals("..")) {
12                 if (list.isEmpty()) continue;
13                 else {
14                     list.removeLast();
15                 }
16             } else {
17                 list.addLast(parts[i]);
18             }
19         }
20         StringBuilder sb = new StringBuilder();
21         int size = list.size();
22         for (int i = 0; i < size; i++) {
23             sb.append("/");
24             sb.append(list.get(i));
25         }
26         return sb.length() == 0? "/": sb.toString();
27     }
28 }