LeetCode_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.

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".

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翻譯：

Code:

/**
 * 
 */
package From61;

import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Deque;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Set;
import java.util.Stack;

/**
 * @author MohnSnow
 * @time 2015年7月1日 上午10:58:22
 * @translate 給定一個Unix風格的路徑，簡化之。使其不改變路徑的結果，但是去掉中間無用的字符。
 *            因爲系統執行的時候也是逐段查看的，因此最直觀的做法就是使用棧來簡化，
 *            當是/..時，出棧；
 *            當是/.時，忽視；
 *            當是//a時進棧/a；
 *            當是其他時才進棧。
 */
public class LeetCode71 {

	/**
	 * @param argsmengdx
	 *            -fnst
	 */
	public static String simplifyPath1(String path) {
		Set<String> isSkip = new HashSet<>(Arrays.asList("", ".", ".."));
		Deque<String> stack = new ArrayDeque<>();
		System.out.println(Arrays.toString(path.split("/")));
		for (String token : path.split("/")) {
			if (token.equals("..") && !stack.isEmpty())
				stack.pop();
			if (isSkip.contains(token))
				continue;
			stack.push(token);
		}
		StringBuilder sb = new StringBuilder();
		while (!stack.isEmpty()) {
			sb.append("/" + stack.pollLast());
		}
		return sb.length() == 0 ? "/" : sb.toString();
	}

	//348msAC----path.split("/")的應用是經典
	public static String simplifyPath2(String path) {
		Deque<String> stack = new ArrayDeque<String>();
		for (String token : path.split("/")) {
			if (token.equals("..") && !stack.isEmpty())
				stack.pop();
			if (token.equals(".") || token.equals("") || token.equals(".."))
				continue;
			stack.push(token);
		}
		System.out.println(stack.toString());
		StringBuilder sb = new StringBuilder();
		while (!stack.isEmpty()) {
			sb.append("/" + stack.pollLast());
		}
		return sb.length() == 0 ? "/" : sb.toString();
	}

	//https://leetcode.com/discuss/22592/java-10-lines-solution-with-stack
	public static String simplifyPath3(String path) {
		Deque<String> stack = new LinkedList<>();
		Set<String> skip = new HashSet<>(Arrays.asList("..", ".", ""));
		for (String dir : path.split("/")) {
			if (dir.equals("..") && !stack.isEmpty())
				stack.pop();
			else if (!skip.contains(dir))
				stack.push(dir);
		}
		String res = "";
		for (String dir : stack)
			res = "/" + dir + res;
		return res.isEmpty() ? "/" : res;
	}

	public static String simplifyPath(String path) {
		StringBuffer result = new StringBuffer();
		int last = 0;
		while (last < path.length()) {
			if (last == (path.length() - 1) && path.charAt(last) == '/') {//  最後一個是/
				break;
			}
			result.append('/');
			while (last + 1 < path.length() && '/' == path.charAt(last + 1)) {//   //--->/
				last++;
			}
			if (last + 1 < path.length() && '.' == path.charAt(last + 1) && '.' == path.charAt(last)) {//  /..----out
				last++;
				last++;
				result.deleteCharAt(result.length() - 1);
				while (result.charAt(result.length() - 1) != '/') {
					result.deleteCharAt(result.length() - 1);
				}
				result.deleteCharAt(result.length() - 1);
				continue;
			}
			if (last + 1 < path.length() && '.' == path.charAt(last + 1)) {//   /.--->忽略此次情況
				last++;
				result.deleteCharAt(result.length() - 1);
				continue;
			}
			while (last < path.length() && '/' != path.charAt(last)) {//  /home--->加入
				result.append(path.charAt(last));
				last++;
			}
		}
		return result.toString();
	}

	public static void main(String[] args) {
		String path = "/home/a/..//a/..";
		System.out.println("path: " + path);
		//System.out.println("simplifyPath: " + simplifyPath(path));
		System.out.println("simplifyPath1: " + simplifyPath1(path));
		System.out.println("simplifyPath2: " + simplifyPath2(path));
		System.out.println("simplifyPath3: " + simplifyPath2(path));
	}
}