71. Simplify Path

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

For example,
path = “/home/”, => “/home”
path = “/a/./b/…/…/c/”, => “/c”
click to show corner cases.

思路：
首先将所有的内容从/中分离出来，然后分别处理。这里我们需要用到堆栈的数据结构。堆栈有很多种实现方式，java中的Stack类LinkedList类都可以实现其功能。我们将读到的路径入栈，根据操作符出栈，最后将栈中剩余的元素组织成String路径返回即可。

class Solution {
    public String simplifyPath(String path) {
        Stack<String> s = new Stack<String>();
        String[] pathDetail = path.split("/");
        for(int i = 0 ; i < pathDetail.length ; i++){
            //弹出上级目录，如果有上级目录
            if (pathDetail[i].equals("..")){
                if (s.isEmpty()){
                    continue;
                }
                s.pop();
            //不进行任何操作
            } else if (pathDetail[i].equals(".") || pathDetail[i].isEmpty()){
                continue;
            } else {
                s.push(pathDetail[i]);
            }
        }
        if (s.isEmpty()){
            return "/";
        }
        StringBuilder result = new StringBuilder();
        do {
            result.insert(0, s.pop());
            result.insert(0, "/");
        } while (!s.isEmpty());
        return result.toString();
    }
}

class Solution {
    public String simplifyPath(String path) {
        if(path == null || path.length() < 1) return path;
        Deque<String> deque = new LinkedList<String>();
        int index = 0;
        StringBuilder ans = new StringBuilder();
        while(index < path.length()) {
            if(path.charAt(index) != '/') {
                StringBuilder dirName = new StringBuilder();
                int dots = 0;
                while(index < path.length() && path.charAt(index) != '/') {
                    dirName.append(path.charAt(index) + "");
                    if(path.charAt(index++) == '.') {
                        dots++;
                    } else dots = 0;
                }
                if(dirName.length() > 2 || dots == 0) deque.add(dirName.toString());
                else if(dots == 2) if(!deque.isEmpty()) deque.removeLast();
            }
            while(index < path.length() && path.charAt(index) == '/') index++;
        }
        if(deque.isEmpty()) return "/";
        while(!deque.isEmpty()) ans.append("/" + deque.removeFirst());
        return ans.toString();
    }
}

class Solution:
    def simplifyPath(self, path: str) -> str:
        lst = path.split('/')
    
        stack = []

        for x in lst:
            if x in ['.','']:
                continue
            if x == '..':
                if stack:
                    stack.pop()
            else:
                stack.append(x)

        return '/' + '/'.join(stack)