137.Single number II (Method 1 sequencing 2STL container map hashing 3-bit operation 4 improved bit operation)

Given an array of integers, every element appears three timesexcept for one. Find the single one.

Note:
Your algorithm should has a linear runtime complexity. Could you implement itwithout using extra memory?

Hidetags

Bit Manipulation

#pragma once#include<iostream> #include <map>using namespace std;//Method 1: Sorting//Method 2:stl container Mapint singleNumber2 ( int a[], int n) {map<int, int> m;map<int, int>::iterator it;for (int i = 0; i < n; i++) {it = M.find (a[i]); I F (It = = M.end ()) m[a[i]] = 1;else//m[a[i]]++;it->second++;} for (int i = 0; i < n; i++) if (m.at (a[i]) = = 1)//at vs. find: At returns the value corresponding to the index, possibly out of bounds return a[i];} Method 3: Bitwise operation.  Consider all binary representations, if we take the sum of all the numbers in the ith position and 3, then only//will have two results 0 or 1 (according to test instructions, 3 0 or 3 1 add remainder is 0). So the result of the remainder is that "single number". int singleNumber3 (int a[], int n) {//int Four bytes 32 bits, per bit and by one digit record int count[32] = {0};//Each element is placed 0int ResU lt = 0;for (int i = 0; i <; i++)//32-bit {for (int j = 0; J < N; j + +)//n number if (A[j] >> 1 & 1)//Note that the operation has only one &amp ;, judge only two Count[i]++;result |= ((Count[i]% 3) << i);//good experience. | : bit or}return result;} Method 4: Improved bit operation. int singleNumber4 (int a[], int n) {int ones = 0, twos = 0, threes = 0;for (int i = 0; i < n; i++) {//note, before each cycle, ones twos t Hrees Disjoint twos |= ones&a[i];//new 2 = original 2 not A[i] + original 2 in A[i]ones ^= a[i];//New 1 = original 1 not a[i] + original non 1 in a[i]//thereupon, new 2 & New 1 = original 2 in A[i]. That is, the intersection part appears 3 times, the intersection part in ones and twos 0,ones twos on the guarantee a[i] after reasonable threes = ones&twos;//Note threes meaning is not a[i] three times the bit but, The newly generated three-bit ones ^= threes;//ones &= ~threes in this layer, but the former is faster twos ^= threes;//twos &= ~threes; Yes, but the former is faster}return ones;} void Main () {int a[] = {1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 5, 5, 5};cout << singleNumber3 (A,) << endl;system ("p Ause ");}

137.Single number II (Method 1 sequencing 2STL container map hashing 3-bit operation 4 improved bit operation)