时间复杂度 O(32n)=O(n),空间复杂度O(1)
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <math.h>
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <limits.h>
#include <string.h>
#include <string>
#include <algorithm>
#include <sstream>
#include <iomanip>
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
using namespace std;
//#define MAXBITNUM 32
//#define MAXNUM 100000
//int bitnumvec[MAXNUM][MAXBITNUM];
int singleNumber(int A[], int n) {
//vector<int*> vec;
if(n==1) return A[0];
const int MAXBITNUM=32;
//int bitnumvec[MAXNUM][MAXBITNUM];
int** bitnumvec=new int*[n];
for(int i=0;i<n;i++)
bitnumvec[i]=new int[MAXBITNUM]();
for(int i=0;i<n;i++)
{
int offset=MAXBITNUM-1;
if(A[i]==-pow(2.0,31))//-2^31
{
bitnumvec[i][0]=1;//, 10000000...000
}
else//others
{
if(A[i]<0&&A[i]>-pow(2.0,31))//negative
{
bitnumvec[i][0]=1;//1 means negative, 0 means positve
A[i]=-A[i];
}
while(A[i]!=0)
{
bitnumvec[i][offset]=A[i]%2;
//bitnum[offset]=A[i]%2;
A[i]=A[i]/2;
offset--;
}
}
//reverse(vec.begin(),vec.end());
//vec.push_back(bitnum);
}
//memset(bitnum,0,sizeof(int)*MAXBITNUM);
int bitnum[MAXBITNUM];
memset(bitnum,0,sizeof(int)*MAXBITNUM);
int x=0;
for(int i=0;i<MAXBITNUM;i++)
{
//if(i==MAXBITNUM-1)
// int y=1;
for(int j=0;j<n;j++)
{
//if(bitnumvec[j][0]==0)
bitnum[i]+=bitnumvec[j][i];
//else if(bitnumvec[j][0]==1)
// bitnum[i]-=bitnumvec[j][i];
}
bitnum[i]=bitnum[i]%3;
if(i>0)
x+=bitnum[i]*pow(2.0,MAXBITNUM-1-i);
}
if(bitnum[0]==1 &&x !=0)
x=-x;
else if(bitnum[0]==1 && x==0)
x=-pow(2.0,31);
//for(int i=0;i<MAXBITNUM;i++)
//int x;
//for(int i=0;i<MAXBITNUM;i++)
for(int i=0;i<n;i++)
delete[] bitnumvec[i];
delete[] bitnumvec;
return x;
}
int main()
{
//int x=-3%2;
int a[]={-2,-2,-2147483648,-2};
cout<<singleNumber(a,4)<<endl;
return 0;
}找唯一不出现三次而出现1次的数子O(n)位运算算法,布布扣,bubuko.com
原文地址：http://blog.csdn.net/richardzrc/article/details/37369351
