[英]Counting special characters as spaces in java
我正在使用JDK 8在MacOS上工作。我想計算給定字符串中特殊字符的數量,但是在給定的代碼中,特殊字符被計為空格。 我該怎么辦?
public static void main(String args[])
{
String str;
int lc=0,uc=0,d=0,s=0,spc=0;
Scanner sc=new Scanner(System.in);
System.out.print("Enter the string:");
str=sc.nextLine();
for(int i=0;i<str.length();i++)
{
if(str.charAt(i)>='a' && str.charAt(i)<='z')
{
lc++;
}
else if(str.charAt(i)>='A' && str.charAt(i)<='Z')
{
uc++;
}
else if(str.charAt(i)>='0' && str.charAt(i)<='9')
{
d++;
}
else if(str.charAt(i)>=32)
{
s++;
}
else if(str.charAt(i)>=33 && str.charAt(i)<=47 || str.charAt(i)==64)
{
spc++;
}
}
System.out.println("number of small characters in "+str+" are:"+lc);
System.out.println("number of CAPITAL characters in "+str+" are:"+uc);
System.out.println("number of digits in "+str+" are:"+d);
System.out.println("number of Spaces in "+str+" are:"+s);
System.out.println("number of Special characters in "+str+" are:"+spc);
}
這是我得到的輸出:
Enter the string:abc23@#$%
number of small characters in abc23@#$% are:3
number of CAPITAL characters in abc23@#$% are:0
number of digits in abc23@#$% are:2
number of Spaces in abc23@#$% are:4
number of Special characters in abc23@#$% are:0
它應該顯示4個特殊字符,而不是4個空格。
用str.charAt(i)==32
更改str.charAt(i)>=32
str.charAt(i)==32
"abc23@#$% "
的輸出將是:
number of small characters in abc23@#$% are:3
number of CAPITAL characters in abc23@#$% are:0
number of digits in abc23@#$% are:2
number of Spaces in abc23@#$% are:1
number of Special characters in abc23@#$% are:4
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