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.

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”.

這個題目重點就是要理解它的意思，如果是一個點 . 那就是當前路徑，不管，如果是兩個點 .. 那就是當前路徑的上一個目錄。這樣我們用棧來表示的話，就是如下所示

path:"/a/./b/../../c/"split:"a",".","b","..","..","c"stack:push(a), push(b), pop(b), pop(a), push(c) --> c

明白這個之后就一目了然了，就是注意返回的時候如果是”/”或者”/../”這種情形就行。

public String simplifyPath(String path) {String res = "";String[] arrs = path.split("/");Stack<String> s = new Stack<String>();for (int i = 0; i < arrs.length; i++) {if (arrs[i].equals("")) {continue;}if (!arrs[i].equals(".") && !arrs[i].equals("..")) {s.push(arrs[i]);}if (arrs[i].equals("..") && !s.isEmpty()) {s.pop();}}if (s.isEmpty())return "/";while (!s.isEmpty())res = "/" + s.pop() + res;return res;}

另一種鏈表的做法

public String simplifyPath1(String path) {String result = "/";String[] stubs = path.split("/+");ArrayList<String> paths = new ArrayList<String>();for (String s : stubs){if(s.equals("..")){if(paths.size() > 0){paths.remove(paths.size() - 1);}}else if (!s.equals(".") && !s.equals("")){paths.add(s);}}for (String s : paths){result += s + "/";}if (result.length() > 1)result = result.substring(0, result.length() - 1); return result;}

總結

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