怎样记算出一个字符串出现的次数````````````````` 比如说:String str1="Request Request Request Request Request Request "; String str2="Request Request ";怎样计算出str2在str1中的次数? 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 public class Test { public static void main(String args[]) { String arg1 = "Request Request Request Request Request Request123123123"; String arg2 = "Request Request"; int count = 0; int current = 0; while ((current = arg1.indexOf(arg2, current)) != -1) { current+=arg2.length(); count++; } System.out.println(count); }} public class Test01 { public static void main(String[] args) { String str1="Request Request Request Request Request Request "; String str2="Request Request "; int count = (str1.length() - str1.replace(str2, "").length()) / str2.length(); System.out.println(count); }}^_^ ^_^ ^_^ String firstString = "Request Request Request Request Request Request ";String replaceString = "Request Request ";String tempString = firstString.replaceAll(replaceString, "");int count = (firstString.length - tempString.length)/replaceString.length; ni...... ni tai huai le ......bi wo kuai......... 这道题要换个我想就不是那么简单了。不过也不难String str = "Request Request Request123123123 Request Request123 Request Request ";String str2 = "Request Request ";这样呢? Why?...My answer is the same as 火龙果. 5555.... v_v 我数学太差了 intcount=(str1.length()-str1.replace(str2,"").length())/str2.length();这句话后的 /str2.length(); 没懂,呵可,谢谢各位指导..... 现成的用apache的common-lang的类库就可以了StringUtils.countMatches(java.lang.String str, java.lang.String sub) 计算匹配个数 For instance:String firstString = "ABCDAB"; //firstString.length = 6String replaceString = "AB"; //replaceString.length =2String tempString = firstString.replaceAll(replaceString, ""); //replaceString.length = 2 int count = (firstString.length - tempString.length)/replaceString.length; // int count = (6-2)/2 = 2Do u know now? 问题是他的requst 可不可以重复计算的???? 呵呵 我给你解释下str1.length()-str1.replace(str2,"").length()str1.replace(str2,"").length()这句是 把str1中只要是str2字符串的内容就换成没有,可不要理解成空格啊这句就是 包含了 str2字符串的 总length数/str2.length();除1个str2的length()数 就是一共包含多少个str2 垃圾回收,一个全局变量在 局部方法中位置Null,当该方法return 时 java,如何使用资源里的图片 热心人来帮忙 急!美萍里的JTabel是怎么做的 abstract是什么意思,下面这句语法怎么理解? String大小写转换的问题 linux下 Java的问题 在线求助 java作业 紧急!!!紧急求助!!! 教教我怎么配置jdk1.3.1_07 关于UI的一个问题,请大家多帮忙,必高分酬谢!! 两行数据重名,查询的时候只能显示其中一个,怎么处理? 大家看看我错在哪了? 刚刚开始学习J2EE,想问问如果J2EE要连接 SQL数据库的问题
public static void main(String args[]) {
String arg1 = "Request Request Request Request Request Request123123123";
String arg2 = "Request Request";
int count = 0;
int current = 0;
while ((current = arg1.indexOf(arg2, current)) != -1) {
current+=arg2.length();
count++;
}
System.out.println(count);
}
}
public static void main(String[] args) {
String str1="Request Request Request Request Request Request ";
String str2="Request Request "; int count = (str1.length() - str1.replace(str2, "").length()) / str2.length();
System.out.println(count);
}
}^_^ ^_^ ^_^
String replaceString = "Request Request ";String tempString = firstString.replaceAll(replaceString, "");int count = (firstString.length - tempString.length)/replaceString.length;
bi wo kuai.........
我想就不是那么简单了。
不过也不难
String str = "Request Request Request123123123 Request Request123 Request Request ";
String str2 = "Request Request ";
这样呢?
我数学太差了 intcount=(str1.length()-str1.replace(str2,"").length())/str2.length();这句话后的 /str2.length(); 没懂,呵可,谢谢各位指导.....
StringUtils.countMatches(java.lang.String str, java.lang.String sub)
计算匹配个数
String firstString = "ABCDAB"; //firstString.length = 6
String replaceString = "AB"; //replaceString.length =2String tempString = firstString.replaceAll(replaceString, ""); //replaceString.length = 2 int count = (firstString.length - tempString.length)/replaceString.length; // int count = (6-2)/2 = 2Do u know now?
呵呵 我给你解释下
str1.length()-str1.replace(str2,"").length()str1.replace(str2,"").length()这句是 把str1中只要是str2字符串的内容就换成没有,可不要理解成空格啊
这句就是 包含了 str2字符串的 总length数
/str2.length();除1个str2的length()数 就是一共包含多少个str2