满四清零的迭代器

可以解决一个整数数组中， 每个数字重复N次，只有一个数字重复了少于N次（比如M次）的问题， 可以构造一个满N清零的迭代器，时间复杂度为O(N)，空间复杂度为O（1）。

	public void counts(int [] A){
		int ones = 0, twos = 0, threes = 0, fours = 0;
		for(int i=0; i<A.length; i++){
			fours = 0;
			fours |= threes & A[i];
			threes |= twos & A[i];
			twos |= ones & A[i];
			ones |= A[i];
			ones &= ~twos;
			ones &= ~threes;
			twos &= ~threes;
			ones &= ~fours;
			threes &= ~fours;
		}
		System.out.println(ones);
		System.out.println(twos);
		System.out.println(threes);
		System.out.println(fours);
	}