如何檢查字符串中的字符是否包含在 JAVA 的括號內？

Question

我有一個要求，我必須從不包含在括號中的字符串中替換多次出現的字符（比如@）。 保證輸入字符串將具有有效的平衡括號。 整個字符串中可以有多對括號。

示例： String str = "今天@如何（你@正在做）@。";

用# 替換“@”符號（括號外）后預期的 output：

你今天怎么樣 ＃。

我可以取出括號內的字符串，進行替換並將括號文本放回去。 但我正在尋找一種更簡單、更清潔的方法。

Answer 1

由於您沒有提及是否可以嵌套以及字符串數據的可能大小。 這是一個可能易於理解的代碼。

如果堆棧為空 => 我們在括號外 => replaceON else OFF

public static void main(String[] args) {
        String data = "How @ are (you @ do(ok@ok)ing) today @.";

        char candidate = '@';
        char replacement = '#';
        boolean replaceFlag = true;

        Stack<Character> stack = new Stack<>();

        StringBuilder newData = new StringBuilder();

        for(char c: data.toCharArray()) {
            if (c == '(') {
                stack.push(c);
                replaceFlag = false;
            } else if(c == ')') {
                stack.pop();
            }

            if (stack.empty()) {
                replaceFlag = true; // we are not inside parenthesis
            }

            if (c == candidate && replaceFlag) c = replacement;

            newData.append(c);
        }

        System.out.println("Old data: "+ data);
        System.out.println("New data: "+ newData);
    }

Answer 2

一個簡單的方法是保持一個“for循環”來讀取字符串的所有字符，例如：

String str = "How @ are (you @ doing) today @.";
        char ab ='';
       for(int i=0;i<str.length();i++)
        {
         ab = str.charAt(i);
         if(ab!='(')
           // check if character is '@' and replace
         else{
           //print as it is; 
            while(ab != ')')
            {//print ab as it is & increment the value of i
            }
           }
           
         }

Answer 3

是的，這正是你需要的..

/**
 *@author jore
 *@version 1.0 2/1/2021
 */
public class replacer
{
  public static void main(String... a)
  {
    /* All these yield correct outpUt */
    String input = "How @ are (you @ doing) today @.";
//    String input = How @ are @ ... @ ... (you @ doing (...@...)..@ ) gghhh @ today @ ( hhh @ uuu (.. @.. ) huu ( hyyy @ bhh @ ) hhh @ ) hhh @..@";
    
    int opened_brackets = 0; //To keep track of opened brackets preventing ..(@..(@..)..#).. and allow complex nesting
    char[] temp_array = input.toCharArray(); //To ease random access in input
    char c; // temp char for testing
    
    for (int i = 0; i < input.length(); i++)
    {
      c = temp_array[i];
      if (c == '(') opened_brackets++; //record opened brackets (reason as explained above
      else if (c == ')') opened_brackets--; //record closure of bracket too...
      if (opened_brackets == 0 && c == '@') temp_array[i] = '#'; // Now replacing is fine if opened brackets are == 0, ot means the character is not in brackets
    }
    
    String output = new String(temp_array); //re-create edited output
    System.out.println(output); //....u know what this does.....¿?
  }
}

Output 1： How # are (you @ doing) today #.

Output 2: # -- replacer How # are #... #... (you @ doing (...@...)..@ ) gghhh # today hhh @ uuu (.. @.. ) huu ( hyyy @ bhh @ ) hhh @ ) hhh #..#

我創建了一個temp_array以允許快速隨機訪問..現在我跟蹤打開的括號以防止替換嵌套括號中的@ ...除了等待)替換..如果打開0個括號..這意味着@放錯了位置（括號外，應立即更換...希望它能解決您的問題