blocks|key|4474989|text|String对象中有一个函数，可以使用常规表达式替换或删除字符串片段。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4474990|public+String+remove(String+base,+String+remove)+{

++return+base.replaceAll(getRegExp(remove),+"");
}

private+String+getRegExp(String+remove)+{

++StringBuilder+regExp+=+new+StringBuilder();

++for+(Character+caracter+:+remove.toCharArray())+{
++++regExp.append("[").append(Character.toLowerCase(caracter))
++++++.append(Character.toUpperCase(caracter)).append("]");
++}

++return+regExp.toString();
}|code-block|syntax|javascript|4474991|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

There is a function within String objects that can replace or remove string snippets using regular expresions.

<pre><code>public String remove(String base, String remove) {

 return base.replaceAll(getRegExp(remove), "");
}

private String getRegExp(String remove) {

 StringBuilder regExp = new StringBuilder();

 for (Character caracter : remove.toCharArray()) {
 regExp.append("[").append(Character.toLowerCase(caracter))
 .append(Character.toUpperCase(caracter)).append("]");
 }

 return regExp.toString();
}
</code></pre>

blocks|key|4475047|text|我的答案是有效的。虽然看起来更简单。期待一个更好的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4475048|public+String+withoutString(String+base,+String+remove)+{
++String+str+=+"";
++for(int+i+=+0;+i+<+base.length()-remove.length()%2B1;i%2B%2B){
++++if(base.substring(i,i%2Bremove.length()).equalsIgnoreCase(remove)){
++++++base+=+base.substring(0,i)%2Bbase.substring(i%2Bremove.length());
++++++i--;
++++}
++}
++return+base;
}|code-block|syntax|javascript|4475049|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My answer works. Seems to be simpler though. looking forward to a better one.

<pre><code>public String withoutString(String base, String remove) {
 String str = "";
 for(int i = 0; i &lt; base.length()-remove.length()+1;i++){
 if(base.substring(i,i+remove.length()).equalsIgnoreCase(remove)){
 base = base.substring(0,i)+base.substring(i+remove.length());
 i--;
 }
 }
 return base;
}
</code></pre>

blocks|key|4474354|text|这对我很有效..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4474355|public+String+withoutString(String+base,+String+remove)+{
++++String+result+=+"";

++++for+(int+i+=+0;+i+<+base.length();+i+%2B=+1)+{
++++++++if+((i+<=+base.length()+-+remove.length())+&&+base.substring(i,+i+%2B+remove.length()).equalsIgnoreCase(remove))+{
++++++++++++i+%2B=+remove.length()+-+1;
++++++++}+else+{
++++++++++++result+%2B=+base.charAt(i);
++++++++}
++++}

++++return+result;
}|code-block|syntax|javascript|4474356|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This works for me...
<pre class="lang-java prettyprint-override"><code>public String withoutString(String base, String remove) {
 String result = &quot;&quot;;

 for (int i = 0; i &lt; base.length(); i += 1) {
 if ((i &lt;= base.length() - remove.length()) &amp;&amp; base.substring(i, i + remove.length()).equalsIgnoreCase(remove)) {
 i += remove.length() - 1;
 } else {
 result += base.charAt(i);
 }
 }

 return result;
}
</code></pre>

blocks|key|4474371|text|public+String+withoutString(String+base,+String+remove)+{
++StringBuilder+sb=+new+StringBuilder();
++sb.append(base);
++++while(sb.toString().toLowerCase().indexOf(remove.toLowerCase())!=-1)
++++sb.replace(sb.toString().toLowerCase().indexOf(remove.toLowerCase()),sb.toString().toLowerCase().indexOf(remove.toLowerCase())%2Bremove.length(),"");
++return+sb.toString();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4474372|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public String withoutString(String base, String remove) {
 StringBuilder sb= new StringBuilder();
 sb.append(base);
 while(sb.toString().toLowerCase().indexOf(remove.toLowerCase())!=-1)
 sb.replace(sb.toString().toLowerCase().indexOf(remove.toLowerCase()),sb.toString().toLowerCase().indexOf(remove.toLowerCase())+remove.length(),"");
 return sb.toString();
}
</code></pre>

blocks|key|4475099|text|public+String+withoutString(String+base,+String+remove)+{
++String+newStr+=+"";
++int+bLength+=+base.length();

++++++int+rLength+=+remove.length();
++++++
++++++for+(int+i+=0;+i+<+bLength;+i%2B%2B){
++++++++if+(i+<=+bLength-rLength+&&+base.substring(i,+i%2BrLength).equalsIgnoreCase(remove){
++++++++++i+%2B=+rLength-1;
++++++++++continue;
++++++++}
++++++++newStr+%2B=+base.charAt(i);
++++++}
++++++return+newStr;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4475100|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public String withoutString(String base, String remove) {
 String newStr = &quot;&quot;;
 int bLength = base.length();

 int rLength = remove.length();
 
 for (int i =0; i &lt; bLength; i++){
 if (i &lt;= bLength-rLength &amp;&amp; base.substring(i, i+rLength).equalsIgnoreCase(remove){
 i += rLength-1;
 continue;
 }
 newStr += base.charAt(i);
 }
 return newStr;
 }
</code></pre>

blocks|key|4475138|text|以下是解决此问题的另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4475139|public+String+withoutString(String+base,+String+remove)+{+
++++return+base.replaceAll("(?i)"+%2B+remove,+"");
}|code-block|syntax|javascript|4475140|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is another way to solve this problem:
<pre><code>public String withoutString(String base, String remove) { 
 return base.replaceAll(&quot;(?i)&quot; + remove, &quot;&quot;);
}
</code></pre>

blocks|key|4475169|text|这是我对这个问题的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4475170|++public+static+String+withoutString(String+str,+String+D)+{
++++String+temp+=+"";
++++for+(int+i+=+0;+i+<+str.length();+i%2B%2B)+{
++++++++temp+%2B=+str.substring(i,+i%2B1);
++++++++if+(temp.length()+>=+D.length()+&&+temp.substring(temp.length()+-+D.length()+,+temp.length()).equalsIgnoreCase(D))+{
++++++++++++temp+=+temp.substring(0,+temp.length()+-+D.length());
++++++++}
++++}
++++return+temp;
}|code-block|syntax|javascript|4475171|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is my answer for this question.
<pre><code> public static String withoutString(String str, String D) {
 String temp = &quot;&quot;;
 for (int i = 0; i &lt; str.length(); i++) {
 temp += str.substring(i, i+1);
 if (temp.length() &gt;= D.length() &amp;&amp; temp.substring(temp.length() - D.length() , temp.length()).equalsIgnoreCase(D)) {
 temp = temp.substring(0, temp.length() - D.length());
 }
 }
 return temp;
}
</code></pre>

Given two strings, <code>base</code> and <code>remove</code>, return a version of the <code>base</code> string where all instances of the <code>remove</code> string have been removed (not case sensitive). 

You may assume that the remove string is length 1 or more. Remove only non-overlapping instances, so with <code>"xxx"</code> removing <code>"xx"</code> leaves <code>"x"</code>. 
For example,

<pre><code>withoutString("Hello there", "llo") → "He there"
withoutString("Hello there", "e") → "Hllo thr"
withoutString("Hello there", "x") → "Hello there"
</code></pre>

But Im failing most of the test cases here in codingbat. Could anybody please help me?

<pre><code>public String withoutString(String base, String remove) {

 String result = "";
 for(int i = 0; i &lt; base.length() - remove.length(); i++){
 if(!(base.substring(i, i + remove.length()).equalsIgnoreCase(remove))){
 result += base.substring(i, i + 1);
 }
 else{
 i = i + remove.length();
 }
 }

 return result;
}
</code></pre>

<a href="https://i.stack.imgur.com/q2XF6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q2XF6.png" alt="enter image description here"></a>

Given two strings, base and remove, return a version of the base string where all instances of the remove string have been removed

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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给定两个字符串base和remove，将返回base字符串的一个版本，其中remove字符串的所有实例都已被删除(不区分大小写)。您可以假设删除字符串的长度为1或更长。只删除不重叠的实例，因此使用"xxx"删除"xx"会留下"x"。例如,withoutString("Hello there", "llo") → "He...

问给定两个字符串base和remove，返回已移除移除字符串的所有实例的基字符串的版本
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问给定两个字符串base和remove，返回已移除移除字符串的所有实例的基字符串的版本EN