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Codility---Brackets

2024-09-18 14:37:10   219人阅读

Task description

A string S consisting of N characters is considered to be properly nestedif 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.

Assume that:

  • N is an integer within the range [0..200,000];
  • string S consists only of the following characters: "(", "{", "[", "]", "}" and/or ")".

Complexity:

  • expected worst-case time complexity is O(N);
  • expected worst-case space complexity is O(N) (not counting the storage required for input arguments).
 
 
Solution
 
Programming language used: Java
Total time used: 14 minutes
Code: 16:26:24 UTC, java, final, score:  100
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// you can also use imports, for example:// import java.util.*;// you can write to stdout for debugging purposes, e.g.// System.out.println("this is a debug message");import java.util.Stack;class Solution {    public int solution(String S) {        // write your code in Java SE 8        if (S.length() % 2 != 0) {            return 0;        }        Character openingBrace = new Character(‘{‘);        Character openingBracket = new Character(‘[‘);        Character openingParen = new Character(‘(‘);        Stack<Character> openingStack = new Stack<Character>();        for (int i = 0; i < S.length(); i++) {            char c = S.charAt(i);            if (c == openingBrace || c == openingBracket || c == openingParen) {                openingStack.push(c);            } else  {                if (i == S.length()-1 && openingStack.size() != 1) {                    return 0;                }                if (openingStack.isEmpty()) {                    return 0;                }                Character openingCharacter = openingStack.pop();                switch (c) {                case ‘}‘:                    if (!openingCharacter.equals(openingBrace)) {                        return 0;                    }                    break;                case ‘]‘:                    if (!openingCharacter.equals(openingBracket)) {                        return 0;                    }                    break;                case ‘)‘:                    if (!openingCharacter.equals(openingParen)) {                        return 0;                    }                    break;                default:                    break;                }            }         }        if (! openingStack.isEmpty()) {            return 0;        }        return 1;    }}


https://codility.com/demo/results/training87ME5J-MVG/

Codility---Brackets


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