Single Number III & 数组中只出现一次的数字

Given an array of numbers 

nums

, in which
exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
For example:
Given 

nums = [1, 2, 1, 3, 2, 5]

, return 

[3,
5]

.
Note:

The order of the result is not important. So in the above example, 

[5, 3]

 is
also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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 Bit Manipulation

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进行两遍异或。

第一遍异或所有元素，得到我们要找的两个元素的异或值。因为我们要找的两个数是不一样的，所以他们异或的结果中肯定有位被置为1.找到其中任意一个被置为1的位，这里我们找最后一个被置为1的位。

第二遍，根据最后一位是否为1，将所有的数分成两组，那么我们要找的两个数必定属于不同的组。然后对这两组分别异或所有元素，就可以得到要找的数了。

注意位操作。

通过将一个数与它的负绝对值进行按位与，可以得到最低位的1。

class Solution
{
public:
vector<int> singleNumber(vector<int>& nums)
{
// Pass 1 :
// Get the XOR of the two numbers we need to find
int diff = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>());
// Get its last set bit
diff &= -diff;

// Pass 2 :
vector<int> rets = {0, 0}; // this vector stores the two numbers we will return
for (int num : nums)
{
if ((num & diff) == 0) // the bit is not set
{
rets[0] ^= num;
}
else // the bit is set
{
rets[1] ^= num;
}
}
return rets;
}
};