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[leetcode]simplify path

admin 11月 11, 2020 0

题目描述

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”.

代码

public class {
  
    public String simplifyPath1(String path) {
        Deque<String> stack = new LinkedList<>();
        Set<String> skip = new HashSet<>(Arrays.asList("..", ".",""));
        for(String dir : path.split("/")){
            if("..".equals(dir) && !stack.isEmpty()) stack.pop();
            else if(!skip.contains(dir)) stack.push(dir);
        }

        // String res = "";
        // for(String dir: stack){
        //     res = "/" + dir + res;
        // }
        StringBuilder res = new StringBuilder();

        while(stack.peekLast() != null){
            res.append("/");
            res.append(stack.pollLast());
        }

        //return res.length() == 0 ? "/" : res;
        return res.length() == 0 ? "/" : res.toString();

    }
}

集合也可以不使用

public class {
  //其实集合也可以不使用
  public String simplifyPath(String path) {
    Deque<String> stack = new LinkedList<>();
    //Set<String> skip = new HashSet<>(Arrays.asList("..", ".",""));
    for(String dir : path.split("/")){
        // if("..".equals(dir) && !stack.isEmpty()) stack.pop();
        // else if(!skip.contains(dir)) stack.push(dir);

        if("..".equals(dir)){
            if(!stack.isEmpty()) stack.pop();
        }else if(!"".equals(dir) && !".".equals(dir)){
            stack.push(dir);
        }
    }

    String res = "";
    for(String dir: stack){
        res = "/" + dir + res;
    }
    
    return res.length() == 0 ? "/" : res;
  }
}