Simplify Path (Java)

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".

/.表示本級目錄，比如/a/.相當於/a， /..表示返回上級目錄，比如/a/../b相當於/b， 這題暗含的條件是多於兩個點的是文件名稱，比如/...a相當於/...a。

這題用兩個棧來保持輸入和輸出順序一致，也可以用一個LinkedList來做，下次寫注意下。

活用split功能，s.split("/")按照斜槓拆分字符串，假如有多個//相連，也會進行拆分，拆分結果是字符數組中存在很多空串。

Source

public class Solution {
    public String simplifyPath(String path) {
    	Stack<String> stack1 = new Stack<String>();
    	String[] str = path.split("/");
    	
    	for(int i = 0; i < str.length; i++){
    		if(str[i].equals(".") || str[i].length() == 0)
    			continue;
    		else if(str[i].equals("..")){
    			if(!stack1.isEmpty())
    				stack1.pop();
    		}
    		else{
    			stack1.push(str[i]);
    		}
    	}
    	
    	StringBuffer a = new StringBuffer();
    	Stack<String> stack2 = new Stack<String>();
    	while(!stack1.isEmpty()){
    		stack2.push(stack1.pop());
    	}
    	while(!stack2.isEmpty()){
    		a.append("/" + stack2.pop());
    	}
    	if(a.length() == 0) a.append("/");
    	
    	return a.toString();
    }
}

Test

    public static void main(String[] args){
    	String path = "/a/./b/../../c/";
    	System.out.println(new Solution().simplifyPath(path));
  
    }