LeetCode——Single Number II

Given an array of integers, every element appears three times except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

给定一个整数数组，除一个外，其它每个元素都出现3次，找出那一个。

可以完全套用上一题的利用哈希表的解法。

另一种解法是：用变量ones   代表二进制1只出现一次的数位，twos  代表二进制1只出现2次的数位，threes  代表二进制1出现3次的数位。当ones和twos中的某一位同时为1时表示二进制1出现3次，此时需要清零。

	public int singleNumber(int[] A) {
		int ones = 0, twos = 0;
		for (int i = 0; i < A.length; i++) {
			twos = twos | (ones & A[i]);
			ones = ones ^ A[i];
			//把出现了3次的位置设置为0,取反
			int threes = ~(ones & twos);
			ones &= threes;
			twos &= threes;
		}
		return ones;
	}

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LeetCode——Single Number II

原文：http://blog.csdn.net/laozhaokun/article/details/34094165