71. Simplify Path

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

Have you met this question in a real interview? 

Yes

Example

"/home/", => "/home"

"/a/./b/../../c/", => "/c"

Challenge 

• 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".

class Solution {
     /* 归下类的话，有四种字符串：
        1. "/"：为目录分隔符，用来分隔两个目录。
        2. "."：当前目录
        3. ".."：上层目录
        4. 其他字符串：目录名

        简化的核心是要找出所有的目录，并且如果遇到".."，需要删除上一个目录。
     */
        public  String simplifyPath(String path) {
        if (path == null || path.length() == 0) return path;

        String[] subPath = path.split("/");
        Stack<String> stack = new Stack<>();
        for (int i = 0; i < subPath.length; i++) {
            String str = subPath[i];
            if (!(str.equals(".") || str.isEmpty())) { // for . and empty string, we do nothing.
                if (str.equals("..")) { // we need to remove one upper directory
                    if (!stack.isEmpty()) {
                        stack.pop();
                    }
                } else {
                    stack.push("/" + str);
                }
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for (String str : stack) {
            sb.append(str);
        }
        return sb.length() == 0 ? "/" : sb.toString();
    }

}

 如果这题改一下

给一个current path, 给一个新的path, 通过cd在当前path执行新的path，问最后的path是什么

这里需要区分新的path是绝对路径还是相对路径，如果新的path是绝对路线，直接不要用当前的path, 如果新的path是相对路径，那么直接把新的path append到当前path，然后再做同样的处理。 

def change_path(current: str, changed: str) -> str:
    if not changed:
        return current
    if changed[0] == "/":
        current = ""

    path = []
    
    for segment in (current + "/" + changed).split("/"):
        if segment == "..":
            if path:
                path.pop()
        elif segment and segment != ".":
            path.append(segment)

    return "/" + "/".join(path)