Leetcode:single number III

Topic:

Given An array of numbers nums , in which exactly-elements appear only once and all the other elements appear exactly Twice. Find the elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5] , return [3, 5] .

Note:

1. The order of the result is not important. The above example, is [5, 3] also correct.
2. Your algorithm should run in linear runtime complexity. Could implement it using only constant space complexity?

Analysis:

Two equal number XOR or result is 0. Therefore, for the first time, the array is scanned, resulting in two separate numbers a, B, or result axorb. Because A and B are unequal, Axorb must not be 0, and bits A and B bits 1 must be different. In any axorb, you can divide the elements of the original array into two sets of XOR results, which are not 0 bits.

Note: N & (-N) is n from low to high, the number corresponding to the first non-0 digit

Reference code:

Class Solution{public:    vector<int> singlenumber (vector<int>& nums)    {    size_t length = Nums.size ();        vector<int> results;    Results.reserve (2);    if (0 = = length) return results;        int axorb = 0;    for (size_t i = 0; i < length; i++)    Axorb ^= nums[i];        int mask = Axorb & (-axorb); Get Axorb from low to high, the first non-0 digits corresponding to the number    int a = 0;    int b = 0;    Axorb bits is 1 bits A and B must be different, using this non-1 bit to divide the original array elements into two sets of XOR results for    (size_t i = 0; i < length; i++)    {    if (Mask & N Ums[i]) a ^= nums[i];    else b ^= nums[i];    }    Results.push_back (a);    Results.push_back (b);    return results;}    ;

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Leetcode:single number III