【LeetCode】71. Simplify Path

内容简介：【LeetCode】71. Simplify Path

问题描述

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

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".

算法

使用 spli('/') 切割字符串，如 /a/./b/../../c/ 切割为： ['a','.','b','..','.','c',''] ，然后就好办了。使用一个栈来存储路径，然后遇到 ['.',''] 就跳过去，遇到 '..' 就将栈的最近元素弹出。

代码

public String simplifyPath(String path) {  
            Stack<String> stack = new Stack<>();
            for(String dir:path.split("/")) {
                if("..".equals(dir)) {
                    if(!stack.isEmpty())
                        stack.pop();
                } else if(".".equals(dir)) {
                    continue;
                } else {
                    if(!"".equals(dir)) {
                        stack.push(dir);
                    }
                }
            }
            String r = "";
            for(String dir:stack) {
                r = r + "/" + dir;
            }
            return "".equals(r)?"/":r;
        }

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