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我正在尝试创建一种方法，该方法将采用两个字符串并返回一个字符串，该字符串已将字符串 1 中括号之间的单词替换为字符串 2 中括号中的单词，但我遇到了我似乎无法理解的问题。举个例子

replaceText("a (simple) programming (example)", "(cool) (problem)")

应该返回

"a cool programming problem"

和

replaceText("a ((nested) example) with (three) replacements (to (handle))", 
            "the replacements are (answer) and (really (two) not three)")

应该返回

"an answer with really (two) not three replacements "

我只能使用循环、基本的 String 方法（.length()、.charAt()）、基本的 StringBuilder 方法和 Character 方法，但我遇到了严重的困难。

现在这是我的代码

public class loopStringAnalysis {

public static String replaceText (String s1, String s2){
StringBuilder newStringBuild = new StringBuilder ();
int count = 0;
int count1 = 0;
for (int i = 0, i1 = 0; i < s1.length() && i1 < s2.length(); i = i + 1){
  if (s1.charAt(i) == '(')
    count = count + 1;
  else if (s1.charAt(i) == ')')
    count = count - 1;
  else if (count == 0)
    newStringBuild.append(s1.charAt(i));
  else if (count != 0){
    while (count1 == 0) {
      if (s2.charAt(i1) == '(')
        count1 = count1 + 1;
      else if (s2.charAt(i1) == ')') 
        count1 = count1 - 1; 
      i1 = i1 + 1;
    }
    while (count1 != 0) {
      if (s2.charAt(i1) == '(')
        count1 = count1 + 1;
      else if (s2.charAt(i1) == ')')
        count1 = count1 - 1;
      else if (count1 != 0)
        newStringBuild.append(s2.charAt(i1));
      i1 = i1 + 1;
    }
    while (count != 0) {
      if (s1.charAt(i) == '(')
        count = count + 1;
      else if (s1.charAt(i) == ')')
        count = count - 1;
      i = i + 1;
    }
  }
}
return newStringBuild.toString();
}

对于第一个示例，这将返回“一个很酷的编程项目”，而对于第二个示例，这将返回“一个答案，实际上是两个而不是三个”。我知道这种方法有问题，但我似乎不知道在哪里。感谢您对修复代码的任何帮助。

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2 回答 2

0

我认为您到目前为止编写的代码过于复杂。使用 String 方法，而不是循环遍历字符串以查找开头和结尾的括号

String.indexOf(char c)

找到括号的索引。这样您就不必遍历第一个字符串。如果您不能使用此方法，请告诉我，我可以尝试提供更多帮助。

于 2013-10-13T21:43:15.333 回答
-1

尝试这个：

public class loopStringAnalysis {

    public static String replaceText(String s1, String s2) {
        StringBuilder newStringBuild = new StringBuilder();
        int count = 0;
        int count1 = 0;
        for (int i = 0, i1 = 0; i < s1.length() && i1 < s2.length(); i = i + 1) {
            if (s1.charAt(i) == '(') {
                count = count + 1;
            } else if (s1.charAt(i) == ')') {
                count = count - 1;
            } else if (count == 0) {
                newStringBuild.append(s1.charAt(i));
            } else if (count != 0) {
                while (count1 == 0) {
                    if (s2.charAt(i1) == '(') {
                        count1 = count1 + 1;
                    } else if (s2.charAt(i1) == ')') {
                        count1 = count1 - 1;
                    }
                    i1 = i1 + 1;
                }
                while (count1 != 0) {
                    if (s2.charAt(i1) == '(') {
                        count1 = count1 + 1;
                    } else if (s2.charAt(i1) == ')') {
                        count1 = count1 - 1;
                    }
                    if (count1 != 0) {
                        newStringBuild.append(s2.charAt(i1));
                    }
                    i1 = i1 + 1;
                }
                while (count != 0) {
                    i = i + 1;
                    if (s1.charAt(i) == '(') {
                        count = count + 1;
                    } else if (s1.charAt(i) == ')') {
                        count = count - 1;
                    }
                }
            }
        }
        return newStringBuild.toString();
    }

    public static void main(String [] args){
        System.out.println(replaceText("a ((nested) example) with (three) replacements (to (handle))",
            "the replacements are (answer) and (really (two) not three)"));
    }
}

让我知道这个是否奏效？

于 2013-10-13T21:47:32.477 回答