LeetCode 71. Simplify Path

原题链接在这里：https://leetcode.com/problems/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"

题解：

首先用string.split 把原有path 按照"/"分开 存入 String 数组strArr中.

"."代表当前层. /a/b/. 还是在b folder中.

从前往后扫strArr. 遇到"." 和 " "直接跳过，遇到正常字符就压入栈中，遇到".."时若stack不是空就pop()一次。

得到最后的stack后一个一个出栈加入到StringBuilder中即可。

Note: 注意corner case 如"///", 此时应该返回"/", 但若是用split拆分是返回一个空的strArr, 所以当stack是空的时候，特殊返回"/".

Time Complexity: O(path.length()). Space: O(path.length()).

AC Java:

class Solution {
    public String simplifyPath(String path) {
        if(path == null || path.length() == 0){
            return path;
        }
        
        //split string into array
        //when encoutering "." or " ", continue
        //when encoutering "..", if stack is not empty, pop stack and return to upper level
        //when encoutering else, push into stack.
        String [] parts = path.trim().split("/");
        Stack<String> stk = new Stack<>();
        for(String part : parts){
            if(part.equals(".")){
                continue;
            }
            
            if(part.equals("..")){
                // Either way, when encountering "..", it needs to skip.
                // Thus it can't combine below if with outter one 
                if(!stk.isEmpty()){
                    stk.pop();
                }
                
                continue;
            }
            
            if(part.length() > 0){
                stk.push(part);
            }
        }
        
        StringBuilder sb = new StringBuilder();
        while(!stk.isEmpty()){
            sb.insert(0, "/"+stk.pop());
        }
        
        // Corner case like "///", it should return "/".
        if(sb.length() == 0){
            sb.insert(0, "/");
        }
        
        return sb.toString();
    }
}