public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.println("Enter a word:");
        String a = in.next();
        System.out.println("Enter another string:");
        String b = in.next();
        if (a.length()!= b.length()){
            System.out.println(b + "is not permutation of the character of " + a);
        }else{
            //如果可以用到工具类就不用自己写排序方法了
            //用Arrays.sort(a.toCharArray())也可以完成
            if(sort(a).equals(sort(b))){
                System.out.println("equality");
            }else{
                System.out.println("inequality");
            }
        }
    }
    
    public static String sort(String str) {
        
        char[] charArray = str.toCharArray();
        
        //这里我就使用了选择排序,当然你也可以用冒泡排序，插入排序
        int flag;
        char temp; 
        for(int i=0;i<charArray.length;i++) {
            flag = i;
            for(int j=i+1;j<charArray.length;j++) {
                if(charArray[flag] > charArray[j]) {
                    flag = j;
                }
            }
            
            if(flag != i) {
                temp = charArray[i];
                charArray[i] = charArray[flag];
                charArray[flag] = temp;
            }
        }
        
        return String.valueOf(charArray);
    }

------解决方案--------------------
第二题很简单的哦~~

Java code
public class StringComparison {

    public static void main(String[] args) {
        String stra = "CCAV没有泷泽萝拉";
        String strb = "泷V萝没泽有AC拉C";
        String str1 = stra.toUpperCase();
        String str2 = strb.toUpperCase();
        int count1 = 0, count2 = 0;
        if (str1.equals(str2)) {
            System.out.println(strb + " 与 " + stra + " 完全相同");
        } else if (str1.length() == str2.length()) {
            for (int i = 0; i < str1.length(); i++) {
                for (int j = 0; j < str2.length(); j++) {
                    if (str1.charAt(i) == str2.charAt(j)) {
                        count2++;
                    }
                }
                for (int k = 0; k < str1.length(); k++) {
                    if (str1.charAt(i) == str1.charAt(k)) {
                        count1++;
                    }
                }
                if (count1 == count2) {
                    count1 = count2 = 0;
                } else {
                    System.out.println(strb + " 不是 " + stra + " 的顺序打乱");//false
                    System.exit(0);
                }
            }
            System.out.println(strb + " 是 " + stra + " 的顺序打乱");//true
        } else {
            System.out.println(strb + " 不是 " + stra + " 的顺序打乱");//false
        }
    }
}

------解决方案--------------------

第一题，用 Integer.bitCount();
for example
Java code
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
System.out.println("1的个数是: " + Integer.bitCount(num));