Problem

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

Example

"/home/", => "/home" //去掉末尾的slash"/a/./b/../../c/", => "/c" //每个"/../"对应：删除一个上层的segment

Challenge

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".

Note

关于challenge的两点：

1. "/../"，这里讨论的有两种情况，空集和"/../"本身。空集加一个if语句返回slash就可以了，"/../"本身要综合Example的例子，pop出上一层元素。

2. Multiple slashes，可以用split()函数去掉所有slash，然后多考虑一个slash中间为空的case。

关于switch语句的一个特点：

我们对case ""和case "."其实都是不做操作，然而，两个case可以都break，但是不能都不做操作。像这样：

case "":case ".":

这样这两个case就都会等价于下一个case：case "..". 就会出错。

Solution

public class Solution {    public String simplifyPath(String path) {        Stack
    
      stack = new Stack
     
      ();        String[] segments = path.split("/");        for (String segment: segments) {            switch(segment) {                case "": break;                case ".":                case "..":                     if (!stack.isEmpty()) {                        stack.pop();                    }                    break;                default: stack.push(segment);            }        }        StringBuilder sb = new StringBuilder();        if (stack.isEmpty()) {//空集的情况            return "/";        }        while (!stack.isEmpty()) {            sb.insert(0, "/"+stack.pop());//Don't miss the slash!        }        return sb.toString();    }}