LeetCode137 Single Number II 独数II

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

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

Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

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题源：here；完整实现：here

思路：

两种方案，一种是该类问题的通用解决方案；一种是特殊优化版。

方案1

我们对int类型的每一个bit位进行记录，如果出现的次数是3的倍数说明应该被移除。

int singleNumber(vector<int>& nums) {
	int res = 0;
	vector<int> single(32, 0);
	for (int i = 0; i < 32; i++){
		for (int j = 0; j < nums.size(); j++){
			if (nums[j] >> i & 1) single[i]++;
		}
		single[i] %= 3;
		res |= single[i] << i;
	}

	return res;
}

方案2

该方案是对方案1的优化。

int singleNumber2(vector<int>& nums){
	int ones = 0, twice = 0, thirds = 0;
	for (int i = 0; i < nums.size(); i++){
		twice |= ones & nums[i];
		ones ^= nums[i];
		thirds = ones & twice;
		ones &= ~(ones&thirds);
		twice &= ~(twice&thirds);
	}

	return ones;
}