룬아님의 취중코딩

Codility 20번 Nesting 본문

개발/알고리즘

Codility 20번 Nesting

룬아님 2019. 9. 9. 17:00

Nesting

 

A string S consisting of N characters is called properly nested if:

  • S is empty;
  • S has the form "(U)" where U is a properly nested string;
  • S has the form "VW" where V and W are properly nested strings.

For example, string "(()(())())" is properly nested but string "())" isn't.

Write a function:

class Solution { public int solution(String S); }

that, given a string S consisting of N characters, returns 1 if string S is properly nested and 0 otherwise.

For example, given S = "(()(())())", the function should return 1 and given S = "())", the function should return 0, as explained above.

Write an efficient algorithm for the following assumptions:

  • N is an integer within the range [0..1,000,000];
  • string S consists only of the characters "(" and/or ")".

    Copyright 2009–2019 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

 

import java.util.Stack;

class Solution {
    public int solution(String S) {
        Stack<Character> sstack = new Stack<Character>();
        
        if(S.equals("")){
            return 1;
        }

        for(int i=0; i<S.length();i++){
            if(S.charAt(i)==')'){
                if(sstack.empty()){
                    return 0;
                }
                if(sstack.pop() != '('){
                    return 0;
                }
            }else{
                sstack.push(S.charAt(i));
            }
        }
        if(!sstack.empty()){
            return 0;
        }
        return 1;
    }
}

https://app.codility.com/demo/results/trainingUAPZAA-6RD/

반응형

'개발 > 알고리즘' 카테고리의 다른 글

Codility 24번 MaxProfit  (0) 2019.09.15
Codility 22번 Dominator  (0) 2019.09.15
Codility 21번 StoneWall  (0) 2019.09.09
Codility 19번 Fish  (0) 2019.09.09
Codility 18번 Brackets  (0) 2019.09.02
Comments