Single Number,single number II

Single numberTotal accepted:103745 Total submissions:218647 difficulty:medium

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

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

(m) Single number II (m) Single number III (m) Missing number (H) Find the Duplicate number

 class   solution { public  :  int  Singlenumber ( Vector<int  >& Nums) { int  res = 0  ;         int  nums_size = Nums.size ();  for  (int  i= 0 ; I<nums_size;i++) {res  ^= Nums[i];     return   res; }};

Next challenges: (m) Single number II (m) Single number III single number IITotal accepted:69333 Total submissions:191725 difficulty:medium

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

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

Count the number of 1 in each bits, this method is suitable for this type of topic: All the numbers in the array appear K, only one appears once

Const intBITS =sizeof(int) *8;classSolution { Public:    intSinglenumber (vector<int>&nums) {        inttimes[bits]={0}; cout<<BITS<<Endl; intNums_size =nums.size ();  for(intI=0; i<nums_size;i++){            intx =Nums[i];  for(intj=0; j<bits;j++){                if((X&GT;&GT;J) &1) {Times[j]++; }            }        }        intres =0;  for(intI=0; i<bits;i++){            if(times[i]%3) {res+=1<<i; }        }        returnRes; }};

Single number IIITotal accepted:18186 Total submissions:44516 difficulty:medium

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

All results differ or are left with a non-0 value, which is the difference or result of only one occurrence of two digits, the number of the first bit=1 in this number, and 1 left to move so many digits into the array of the splitter.

Const intBITS =sizeof(int)*8;classSolution { Public: Vector<int> Singlenumber (vector<int>&nums) {        intNums_size =nums.size (); intn =0;  for(intI=0; i<nums_size;i++) {n^=Nums[i]; }        intSeprator =0;  for(intI=0; i<bits;i++){            if((n>>i) &1) {Seprator=1<<i;  Break; }} vector<int>Res; intres1=0, res2=0;  for(intI=0; i<nums_size;i++){            if((Seprator &Nums[i])) {Res1^=Nums[i]; }Else{res2^=Nums[i];        }} res.push_back (RES1);        Res.push_back (Res2); returnRes; }};

Next challenges: (E) Number of 1 Bits (m) Bitwise and of Numbers Range (m) Maximum Product of Word Lengths

Single Number,single number II