Question

如何从字符串中删除单词？例如，如果我有以下输入：

Cancer is a fatal disease

Cancer is afataldisease

Cancer is a fataldisease

Cancer is afataldisease

Cancer is a     fatal     disease

Cancer is     afataldisease

Cancer is a     fataldisease

Cancer is     afataldisease

输出应该是：

Cancer is a disease(for first 4 cases)

剩下的：

Cancer is a           disease

Cancer is     a disease

Cancer is a      disease

Cancer is     a disease

我有前4个案例的输出，但不是上面的4个案例。在最后4个案例中，删除单词后留下了额外的空格。

String deleteWord(String sentence,String... words)
{
    for(String s:words)
    {
        sentence=sentence.replaceAll(s," ");
    }
    return sentence;
}

Answer 1

你应该用空格替换致命一词。之后检查每个句子中的多个常规空格。如果您有多个常规空格，请将其替换为单个空格。

public class ReplaceWord {
    public static void main(String[] args) {
        String[] str = {
"Cancer is a fatal disease",

"Cancer is afataldisease",

"Cancer is a fataldisease",

"Cancer is afataldisease",

"Cancer is a     fatal     disease",

"Cancer is     afataldisease",

"Cancer is a     fataldisease",

"Cancer is     afataldisease"};



    for(String string: str) {
        string = string.replace("fatal", " ").replaceAll("\\s+", " ");
        System.out.println(string);

    }



    }}

正则表达式＆＃34; \\ s +＆＃34; 将检查多个常规空格。如果找到，它将被替换为单个空格。

Answer 2

我修改了我的代码以匹配你的查询。我最初误解了你对非连续空间的需求，现在这段代码保存了你单词的每个单一标记的格式（它的左外，右索引和一个整数来告诉天气空间是连续）在整数ArrayList中并根据该数据修改句子。所以现在它在开始或结束时重复工作。即使你提到的两种句子都在一个句子中。

String sentence= " fatal Cancer is a fatal diseasefatal";//one example
String word="fatal";

int spaces=0;
int left=-1;
int right=0;
int nonspacechar=0;//leftmost nonspacechar
int difference=0;//difference between the outermost indexes
int consecspaces=0;

ArrayList<Integer> format = new ArrayList();//declare arrayList

for(int i =0;i<sentence.length();i++){


if(i==sentence.indexOf(word,right)){//to get the left index after the word's right index

    if(nonspacechar==0){
        left=nonspacechar;
    }else
left=nonspacechar+1;

format.add(left);//adding the left index to the arraylist
if(sentence.length()-i==word.length()){//if the word is at the end
   right=i+word.length();
format.add(right);


     format.add(2); 
break;
}


i=i+word.length()-1;

continue;
}

if(sentence.charAt(i)!=' '){
    if(left>=0){//to get the right non space index when left is set
right=i;
format.add(right);//add right outermost index

if(i==sentence.length()-1&&sentence.charAt(i)=='.'){
        format.add(2);
    }else

if(spaces>1){
    format.add(1);

}else if(nonspacechar==0){
       format.add(2); 
}else
     format.add(0);



        left=-1;//reset left

}
nonspacechar=i;
    spaces=0;
}else if(left>=0&&spaces==1){
spaces=0;
}else
spaces++; 
}


int k=0;
StringBuilder newSentence=new StringBuilder(sentence);//declare the newsentence Stringbuilder to be replaced acording to indexes

while(k<format.size()){

    left=format.get(k);
    right=format.get(k+1);


    if(k>0){
        if(consecspaces==1){//check last consecuative space condition

        left=left-difference;
        right=right-difference; 
        }else{
          left=left-(difference);//subtract last difference
        right=right-(difference);  
        }
    }

     consecspaces=format.get(k+2);




if(consecspaces==1){//check if consecuativespaces exist
  difference=difference+word.length();//adding word on difference
 newSentence=newSentence.replace(newSentence.indexOf(word,left),newSentence.indexOf(word,left)+word.length(),"");

}else if(consecspaces==2){ 
      difference=difference+(right-left);
      newSentence=newSentence.replace(left,right,"");
}else{
     difference=difference+(right-left)-1; //adding word on difference
     newSentence=newSentence.replace(left,right," ");
}
k=k+3;
}

sentence=new String(newSentence);//get the new sentence
    System.out.print(sentence);

}


}

当然要根据索引替换字符串我必须使用StringBuilder也不要忘记导入Arraylist： -

import java.util.ArrayList;

注意：如果这个词是连续重复的，那么这个算法不起作用，我肯定不是语法：）。

Answer 3

更换单词＆＃34;致命＆＃34;单个空格，

尝试使用单个空格（＆＃34;＆＃34;）拆分整个字符串，然后在迭代返回的字符串数组时删除单个空格元素，并将单词空间合并回来。因此，这些不需要的空间将被删除。

如何从Java中删除字符串中的单词？

3 个答案: