# Single Number II Solutions in C++

Number 137

Difficulty Medium

Acceptance 52.5%

Link LeetCode

Other languages —

## Solutions

### C++ solution by haoel/leetcode

// Source : https://oj.leetcode.com/problems/single-number-ii/// Author : Hao Chen// Date : 2014-06-17class Solution {public:Solution(){srand(time(0));}//random invokerint singleNumber(int A[], int n) {if (rand()%2){return singleNumber_1(A, n);}return singleNumber_2(A, n);}/** This solution is clear & straightforward implementation.** We use an array of 32 length(e.g. count[32]) to count the the bits for all of numbers.** Because the same number appear 3 times, which means the sum of i-th bits for all numbers should be 3 times.** In other word, the sum of the i-th bits mod 3, it must be 0 or 1. 1 means that is the single number bit.** This solution can be easy to extend to "every element appears k times except for one."**/int singleNumber_1(int A[], int n) {int count[32] = {0};int result = 0;for (int i = 0; i < 32; i++) {for (int j = 0; j < n; j++) {if ((A[j] >> i) & 1) {count[i]++;}}result |= ((count[i] % 3) << i);}return result;}/** The following solution is popular solution on Internet, but it looks it's not easy to understand.** Actually, it just optimizes the above soultion.** Let's see how it improve the above.** We use three bitmasks,* 1) `ones` as a bitmask which represents the i-th bit had appeared once.* 2) `twos` as a bitmask which represents the i-th bit had appeared twice.* 3) `threes` as a bit mask which represents the i-th bit had appeared three times.** When the i-th bit had appeared for the third time, clear the i-th bit of both `ones` and `twos` to 0.* The final answer will be the value of `ones`**/int singleNumber_2(int A[], int n) {int ones = 0, twos = 0, threes = 0;for (int i = 0; i < n; i++) {// `ones & A[i]` the result is the bitmask which the bits appeared twicetwos |= ones & A[i];// XOR means remove the bit which appeared twice int `ones`ones ^= A[i];// count the `three`threes = ones & twos;// clear the `ones` and `twos` if the i-th bit had appeared three times.ones &= ~threes;twos &= ~threes;}return ones;}};

### C++ solution by pezy/LeetCode

class Solution {public:int singleNumber(int A[], int n) {int low = 0;for (int i = 0, high = 0; i != n; ++i){low ^= A[i] & ~high;high ^= A[i] & ~low;}return low;}};