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

public class Solution {
    public String simplifyPath(String path) {
        if (path == null || path.length() == 0 || path.equals("")) {
            return "";
        }
        LinkedList<String> list = new LinkedList<String>();
        String[] parts = path.split("/");

        for (int i = 0; i < parts.length; i++) {
            if (parts[i].equals("") || parts[i].equals(".")) continue;
            else if (parts[i].equals("..")) {
                if (list.isEmpty()) continue;
                else {
                    list.removeLast();
                }
            } else {
                list.addLast(parts[i]);
            }
        }
        StringBuilder sb = new StringBuilder();
        int size = list.size();
        for (int i = 0; i < size; i++) {
            sb.append("/");
            sb.append(list.get(i));
        }
        return sb.length() == 0? "/": sb.toString();
    }
}

更多讨论：

https://discuss.leetcode.com/category/79/simplify-path