讲两个int 数组找出重复的数字 用最少的循环

        int a[] = {1,3};
        int b[] = {1,3,5};
        int size = a.length>b.length ?a.length:b.length;
        int valueA =0 ;
        int valueB =0;
        for(int i =0;i<size;i++){
            int tempA =1;
            int tempB =1;
            if(i<a.length){
                valueA += tempA << a[i];
                valueB += tempB << b[i];
            } else {
                valueB += tempB << b[i];    
            }
            
        }
        System.out.println(Integer.toBinaryString(valueA & valueB));
	1010
	3210
  int b[] = {1,3,8,5};
        int a[] = {1,3,4,5,8};
        boolean flag = a.length>b.length ?true:false;
        int size = a.length>b.length ?a.length:b.length;
        Set<Integer> setInt = new HashSet<Integer>();
        for (int i=0; i<size; i++){
            if(!flag && i<a.length){
                setInt.add(a[i]);
            }
            if(flag && i<b.length){
                setInt.add(b[i]);
            }
            if(!flag &&setInt.contains(b[i])){
                System.out.println(b[i]);
            }
            if(flag &&setInt.contains(a[i])){
                System.out.println(a[i]);
            }
        }
    }


原文地址:https://www.cnblogs.com/root429/p/9251356.html