[LeetCode] 71. Simplify Path

轉自 LeetCode

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".

<Solution>

這題是要去找出最後最精簡的 path

可以看到 input 是用 '/ ' 來做切分

因此，可以使用 stringstream 來幫忙做這件事

利用 stringstream 將兩個 / 之間的字串給取出來後，可能有以下情況

• 空字串 : 例如 "//"

• "." : 代表當下位置，沒有移動

• ".." : 到上一層

• 其餘就是檔案或是目錄名稱

這邊附上一些測資

"/"
"/../"
"/home//file"
"/home/"
"/a/./b/../../c/"
"/..."
"/."
"/..hidden"
"/.hidden"
"/home/foo/.ssh/authorized_keys/"

code 如下
c++

	class Solution {
	public:
	string simplifyPath(string path) {
	string tmpStr;
	stringstream ss(path);
	vector<string> token;
	while(getline(ss, tmpStr, '/')) {
	if(tmpStr == "" || tmpStr == ".") {
	continue;
	}
	if(tmpStr == ".." && !token.empty()) {
	token.pop_back();
	}
	else if(tmpStr != "..") {
	token.push_back(tmpStr);
	}
	}
	//> combine string in reverse order
	string ans;
	for(auto s : token) {
	ans += "/" + s;
	}
	return (token.empty()) ? "/" : ans;
	}
	};

view raw simplifyPath.cpp hosted with ❤ by GitHub

kotlin

	class Solution {
	fun simplifyPath(path: String): String {
	val paths = path.split("/")
	val tokens = mutableListOf<String>()
	loop@ for(p in paths) {
	when {
	p.isBlank() || p == "." -> continue@loop
	p == ".." && tokens.isNotEmpty() -> tokens.removeAt(tokens.lastIndex)
	p != ".." -> tokens.add(p)
	}
	}
	return if (tokens.isEmpty()) {
	"/"
	} else {
	val sb = StringBuilder()
	for(t in tokens) {
	sb.append("/$t")
	}
	sb.toString()
	}
	}
	}

view raw simplify_path.kt hosted with ❤ by GitHub