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". /* 精简目录 note:遇到.则表示当前,遇到..则表示回到父目录,根目录的父目录还是/ 栈处理 Input: "/abc/..." Output: "/abc/..." */ public class Solution { public String simplifyPath(String path) { Stack<String> stack = new Stack<String>(); String[] split = path.split("/"); for(String str : split) { if(!str.equals(".") && !str.isEmpty()) { if(str.equals("..")) { if(!stack.isEmpty()) stack.pop(); } else { stack.push(str); } } } StringBuilder sb = new StringBuilder(); if(stack.isEmpty()) sb.insert(0, "/"); else { while(!stack.isEmpty()) { sb.insert(0, stack.pop()); sb.insert(0, "/"); } } return sb.toString(); } }
路径简化算法
本文介绍了一种简化Unix风格绝对路径的方法,通过使用栈来处理遇到的符号'.'和'..',并给出了具体的实现代码。
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