[Codility] Stacks and Queues. Brackets

https://app.codility.com/programmers/lessons/7-stacks_and_queues/brackets/

A string S consisting of N characters is considered to be properly nested if any of the following conditions is true:
S is empty;S has the form "(U)" or "[U]" or "{U}" where U is a properly nested string;S has the form "VW" where V and W are properly nested strings.
For example, the string "{[()()]}" is properly nested but "([)()]" is not.
Write a function:
class Solution { public int solution(String S); }
that, given a string S consisting of N characters, returns 1 if 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..200,000];string S consists only of the following characters: "(", "{", "[", "]", "}" and/or ")".

 

쉬운 문제입니다. 

스택을 이용해서 처리 할 수 있습니다.

 

스택이 비어있으면 push를, 그리고 쌍일 경우에 pop을 합니다. 그리고 모든 스택이 비어있으면 1을 그렇지 않으면 0을 리턴합니다.

매우 간단한 구조입니다.

 

아래는 소스코드 입니다.

 

public class Brackets {
    public int solution(String S) {
        Stack<Character> stack = new Stack<>();

        for (int i = 0; i < S.length(); i++) {
            if (stack.isEmpty()) {
                stack.push(S.charAt(i));
                continue;
            }

            if ((stack.peek() == '(' && S.charAt(i) == ')')
                    || (stack.peek() == '[' && S.charAt(i) == ']')
                    || (stack.peek() == '{' && S.charAt(i) == '}')) {
                stack.pop();
                continue;
            }

            stack.push(S.charAt(i));
        }

        return stack.isEmpty() ? 1 : 0;
    }
}