Leetcode - Simplify Path

最新推荐文章于 2020-09-18 20:20:43 发布

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最新推荐文章于 2020-09-18 20:20:43 发布

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原文链接：http://www.cnblogs.com/shuaih/p/7435861.html

本文介绍了一个算法用于简化Unix风格文件路径。通过解析路径字符串并利用栈来存储有效的目录部分，该算法能有效处理.和..等特殊符号，最终返回规范化的最简路径。

Question:

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

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

===========================================================================================================

We need to know:

"." current working directory /a /. / the command does not direct to any other directories. Basically it does nothing
".." jump one level back / c / a / .. / the current working directory is "a", if we do ".." after "a", then we jump one level back to "c"
"/" this is the divider of commands, which means "///" no matter how many grouped together, it is just " / ".
other than the previous three mentioned, they are all directory names. Those directories are the ones we need to store

Because reading the commands starts from left to right, it makes sense that we will need to loop through the command line and evaluate the meaning of each command. As we will need print out the meaningful path, we need to store each meaningful command, so we use stack. It makes becuase when we have "..", we need to know previous directory and delete it. Stack will allow us to do it this way easily.

 1 class Solution {
 2     public String simplifyPath(String path) {
 3         String[] strs = path.split("/");
 4         //create a stack to store meaningful command
 5         Stack<String> stack = new Stack<>();
 6         for (int i = 0; i < strs.length; i++) {
 7             //jump one directory back
 8             if (strs[i].equals("..")) {
 9                 if (!stack.isEmpty()) {
10                     stack.pop();
11                 }
12             } else if ((!strs[i].equals(".")) && (!strs[i].equals(""))) {
13                 stack.push(strs[i]);
14             }
15             //if the command is . or empty, then the command is not meaningful, so we will skep the execution
16         }
17         //print out the path
18         path = "";
19         while (!stack.isEmpty()) {
20             path = "/" + stack.pop() + path;
21         }
22         if (path.length() > 0) {
23             return path;
24         }
25         return "/";
26     }
27 }

转载于:https://www.cnblogs.com/shuaih/p/7435861.html