考虑括号的Java StringTokenizer

Question

我正在创建一个程序，它对 boolean 逻辑表达式进行标记并返回标记的字符串数组。以下是我的代码:

public static String[] tokenize(String s)
{
    String delims = "+";
    StringTokenizer st = new StringTokenizer(s, delims);
    String[] tokens = new String[st.countTokens()];

    int i=0;
    while (st.hasMoreElements()) {
        tokens[i++] = st.nextElement().toString();
    }

    return tokens;
}

例如，我有以下字符串作为输入:

A+B+(C+D+(A+B))+(B+C)

使用我的代码，它只会生成以下标记:

A
B
(C
D
(A
B))
(B
C)

是否有可能(使用相同的代码结构)提出这些标记？如果不是，它如何编码？

A
B
(C+D+(A+B))
(B+C)

Answer 1

    ArrayList<String> tokens = new ArrayList<String>();
    String current = "";
    int counter = 0;
    for(int i = 0 ; i < input.length(); i++)
    {
        char c = input.charAt(i);
        if(counter==0 && c=='+')
        {
            tokens.add(current);
            current = "";
            continue;
        }
        else if(c=='(')
        {
            counter++;
        }
        else if(c==')')
        {
            counter--;
        }
        current += c;
    }
    tokens.add(current);

这是我评论的解决方案:

You could just loop through 1 character at a time, when you reach a + while not in a parenthesis, save the characters read up to there, and start a new set. The way to track if you're in a a set of parentheses is with a counter. When you hit a open parenthesis, you increment a counter by 1, when you hit a close parenthesis you decrement the counter by 1. If the counter > 0 then you're in parentheses. Oh, and if counter ever goes negative, the string is invalid, and if counter is not 0 at the end, then the string is also invalid.

您可以对计数器进行检查并返回 false 或其他东西，以表明它是一个无效的字符串。如果您知道该字符串是有效的，那么它可以按原样工作。您可以使用 tokens.toArray()

从 ArrayList 获取数组