java如何检查字符串是否平衡？

Question

我想测试输入字符串是否平衡。如果有一个匹配的开始和结束括号、括号或大括号，它将是平衡的

example:
{} balanced
() balanced
[] balanced
If S is balanced so is (S)
If S and T are balanced so is ST


public static boolean isBalanced(String in)
{
    Stack st = new Stack();

    for(char chr : in.toCharArray())
    {
        if(chr == '{')
            st.push(chr);

    }

    return false;
}

我在选择做什么时遇到了问题。我是否应该将每个开始或结束的括号、括号或大括号放在一个堆栈中，然后将它们弹出？如果我把它们拿出来，那对我有什么帮助

Answer 1

1）对于每个打开的括号：{ [ (将其推到堆栈中

2）对于每个结束括号：} ] )从堆栈中弹出并检查括号的类型是否匹配。如果不返回false

即字符串中的当前符号是}，如果从堆栈中弹出的是{中的任何其他符号，则立即返回false

3）如果行尾和堆栈不是空的，则返回false，否则返回true

Answer 2

下面是一个Java代码示例，用于检测字符串是否平衡

http://introcs.cs.princeton.edu/java/43stack/Parentheses.java.html

我们的想法是——

• 对于每个打开的大括号( [ {，将其推到堆栈上
• 对于右大括号) ] }，请尝试从堆栈中弹出匹配的左大括号( [ }。若你们找不到匹配的大括号，那个么字符串是不平衡的
• 如果处理完完整的字符串后，堆栈为空，则字符串处于平衡状态。否则字符串不平衡

Answer 3

是的，堆栈是任务的合适选择，也可以使用递归函数。如果你使用一个堆栈，那么你的想法是推堆栈上的每个打开的括号，当你遇到一个关闭的括号时，你检查堆栈的顶部是否匹配它。如果匹配，则将其弹出，如果不匹配，则为错误。完成后，堆栈应为空

import java.util.Stack;
public class Balanced {
    public static boolean isBalanced(String in)
    {
        Stack<Character> st = new Stack<Character>();

        for(char chr : in.toCharArray())
        {
            switch(chr) {

                case '{':
                case '(':
                case '[':
                    st.push(chr);
                    break;

                case ']':
                    if(st.isEmpty() || st.pop() != '[') 
                        return false;
                    break;
                case ')':
                    if(st.isEmpty() || st.pop() != '(')
                        return false;
                    break;
                case '}':
                    if(st.isEmpty() || st.pop() != '{')
                        return false;
                    break;
            }
        }
        return st.isEmpty();
    }
    public static void main(String args[]) {
        if(args.length != 0) {
            if(isBalanced(args[0]))
                System.out.println(args[0] + " is balanced");
            else
                System.out.println(args[0] + " is not balanced");
        }
    }
}

Answer 4

粗略地说，如果它是平衡的，这意味着你的堆栈应该是空的

为此，在解析}时需要弹出堆栈

额外的要求是检查}前面是否有{或者弹出的字符是否是{

Answer 5

import java.util.Stack;

public class SyntaxChecker {

    /**
     * This enum represents all types of open brackets. If we have a new type then
     * just add it to this list with the corresponding closed bracket in the other
     * ClosedBracketTypes enum  
     * @author AnishBivalkar
     *
     */
    private enum OpenBracketTypes {
        LEFT_ROUND_BRACKET('('),
        LEFT_SQUARE_BRACKET('['),
        LEFT_CURLY_BRACKET('{');

        char ch;

        // Constructs the given bracket type
        OpenBracketTypes(char ch) {
            this.ch = ch;
        }

        // Getter for the type of bracket
        public final char getBracket() {
            return ch;
        }


        /**
         * This method checks if the current character is of type OpenBrackets
         * @param name
         * @return True if the current character is of type OpenBrackets, false otherwise
         */
        public static boolean contains(final char name) {
            for (OpenBracketTypes type : OpenBracketTypes.values()) {
                if (type.getBracket() == name) {
                    return true;
                }
            }

            return false;
        }
    }

    /**
     * This enum represents all types of Closed brackets. If we have a new type then
     * just add it to this list with the corresponding open bracket in the other
     * OpenBracketTypes enum    
     * @author AnishBivalkar
     *
     */
    private enum CloseBracketTypes {
        RIGHT_ROUND_BRACKET(')'),
        RIGHT_SQUARE_BRACKET(']'),
        RIGHT_CURLY_BRACKET('}');

        char ch;
        CloseBracketTypes(char ch) {
            this.ch = ch;
        }

        private char getBracket() {
            return ch;
        }

        /**
         * This method checks if a given bracket type is a closing bracket and if it correctly
         * completes the opening bracket
         * @param bracket
         * @param brackets
         * @return
         */
        public static boolean isBracketMatching(char bracket, Stack<Character> brackets) {
            // If the current stack is empty and we encountered a closing bracket then this is
            // an incorrect syntax
            if (brackets.isEmpty()) {
                return false;
            } else {
                if (bracket == CloseBracketTypes.RIGHT_ROUND_BRACKET.getBracket()) {
                    if (brackets.peek() == OpenBracketTypes.LEFT_ROUND_BRACKET.getBracket()) {
                        return true;
                    }
                } else if (bracket == CloseBracketTypes.RIGHT_SQUARE_BRACKET.ch) {
                    if (brackets.peek() == OpenBracketTypes.LEFT_SQUARE_BRACKET.getBracket()) {
                        return true;
                    }
                } else if (bracket == CloseBracketTypes.RIGHT_CURLY_BRACKET.ch) {
                    if (brackets.peek() == OpenBracketTypes.LEFT_CURLY_BRACKET.getBracket()) {
                        return true;
                    }
                }

                return false;
            }
        }

        /**
         * This method checks if the current character is of type ClosedBrackets
         * @param name
         * @return true if the current character is of type ClosedBrackets, false otherwise
         */
        public static boolean contains(final char name) {
            for (CloseBracketTypes type : CloseBracketTypes.values()) {
                if (type.getBracket() == name) {
                    return true;
                }
            }

            return false;
        }
    }


    /**
     * This method check the syntax for brackets. There should always exist a
     * corresponding closing bracket for a open bracket of same type.
     * 
     * It runs in O(N) time with O(N) worst case space complexity for the stack
     * @param sentence The string whose syntax is to be checked
     * @return         True if the syntax of the given string is correct, false otherwise
     */
    public static boolean matchBrackets(String sentence) {
        boolean bracketsMatched = true;

        // Check if sentence is null
        if (sentence == null) {
            throw new IllegalArgumentException("Input cannot be null");
        }

        // Empty string has correct syntax
        if (sentence.isEmpty()) {
            return bracketsMatched;
        } else {
            Stack<Character> brackets = new Stack<Character>();
            char[] letters = sentence.toCharArray();

            for (char letter : letters) {

                // If the letter is a type of open bracket then push it 
                // in stack else if the letter is a type of closing bracket 
                // then pop it from the stack
                if (OpenBracketTypes.contains(letter)) {
                    brackets.push(letter);
                } else if (CloseBracketTypes.contains(letter)) {
                    if (!CloseBracketTypes.isBracketMatching(letter, brackets)) {
                        return false;
                    } else {
                        brackets.pop();
                    }
                }
            }

            // If the stack is not empty then the syntax is incorrect
            if (!brackets.isEmpty()) {
                bracketsMatched = false;
            }
        }

        return bracketsMatched;
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        String words = "[[][][]Anfield[[]([])[]]becons()]";
        boolean isSyntaxCorrect = SyntaxChecker.matchBrackets(words);

        if (isSyntaxCorrect) {
            System.out.println("The syntax is correct");
        } else {
            System.out.println("Incorrect syntax");
        }
    }
}

对此的任何反馈都是非常欢迎的。如果你发现有什么错误或无用的东西，请批评。我只是在努力学习

Answer 6

我编写这段代码是为了解决这个问题，在每个括号类型中只使用一个整数（或者可能是一个字节）变量

public boolean checkWithIntegers(String input) {

    int brackets = 0;

    for (char c: input.toCharArray()) {

        switch (c) {

            case '(':

                brackets++;

                break;

            case ')':

                if (brackets == 0)
                    return false;

                brackets--;

                break;

            default:

                break;
        }
    }


    return brackets == 0;
}

public static void main(String... args) {

    Borrar b = new Borrar();
    System.out.println( b.checkWithIntegers("") );
    System.out.println( b.checkWithIntegers("(") );
    System.out.println( b.checkWithIntegers(")") );
    System.out.println( b.checkWithIntegers(")(") );
    System.out.println( b.checkWithIntegers("()") );

}

OBS

1. 我假设一个空字符串是平衡的。这可以通过之前的检查来改变
2. 我只使用了一种类型的括号示例，但只需添加另一个case开关即可快速添加其他类型

希望这有帮助。 干杯