Single Number 系列

Single Number

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

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

class Solution {public:    int singleNumber(vector<int>& nums) {        int diff = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>());        return diff;    }};

Single Number II

Given an 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?

class Solution {public:    int singleNumber(vector<int>& nums) {        int one = 0, two = 0, three = 0;        for (int i = 0; i < nums.size(); ++i) {            two |= one & nums[i];            one ^= nums[i];            three = one & two;            one &= ~three;            two &= ~three;        }        return one;    }};

Single Number III

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?

class Solution {public:    vector<int> singleNumber(vector<int>& nums) {        int diff = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>());        diff &=-diff;        vector<int> result(2,0);        for(auto num : nums){            if(num&diff) result[0] ^=num;            else result[1]^=num;        }        return result;    }};