面试题：给40亿个不重复的无符号整数，没排过序，给一个无符号整数如何快速判断这个数是否在这40亿个数中

//**********************位图***********************

//腾讯面试题：

//           给40亿个不重复的无符号整数，没排过序，给一个无符号整数如何

//           快速判断这个数是否在这40亿个数中

//解题思路：

//（1）直接存进内存

//40亿个无符号数，如果放到内存，就需要开辟4*4G=16G的空间，因为数的范围不确定，所以

//开辟尽可能大的空间来存放，需42亿9千万×4字节的空间

//（2）位图   

//位图：在一块内存区域的每个比特存0或1，表示其对应的元素不存在或者存在

//位图优点： 速度快，内存空间占用小，能表示大范围的数据

//比较：用位图的话，需（42亿9千万×4字节）/32字节=大约500M的空间内存就可以

//      把40亿个数全部放进内存

//按位与&：都为真时才为真，即都为1时才为1

//按位或|：都为假时才为假，即都为0时才为0

//按位异或^：两个数不同时为1

//左移<<是往高位移     如在一个从0到31的比特位中，让1<<2,则比特位变为00100000  00000000   00000000  00000000

//右移>>是往低位移   x>>1等于x/2   x>>5等于x/32

#include<iostream>

#include<vector>

usingnamespacestd;

classBitMap

{

public:

                BitMap(size_trange)

                                :size(0)

                {

                                arr.resize((range>> 5) + 1);

                }

                //把某个数对应的比特位标记为1表示这个数存在

                boolSet(size_tdata)

                {

                                //找data对应的比特位

                                size_tindex =data>> 5;     

                                size_tnum =data% 32;

                                //arr是vector类型的，会被自动初始化为0

                                //1<<num时只有一位为1，其余都为0，当数据对应的比特位为0时进入循环，防止对一个数据进行多次标记，使size和实际数据不符

                                if(!(arr[index] & (1 << num)))     

                                {

                                                arr[index] |= (1 << num);     //arr中的值与(1 << num)保证了已经被标记为1的地方不变

                                                ++size;                      

                                                returntrue;

                                }

                                else

                                {

                                                returnfalse;

                                }

                }

                boolReset(size_tdata)

                {

                                //找data对应的比特位

                                size_tindex =data>> 5;

                                size_tnum =data% 32;

                                //当数据对应的比特位为1时进入循环，防止多次Reset，size在一个数上面多次--

                                if(arr[index] & (1 << num))    

                                {

                                                arr[index] &= (~(1 << num));   

                                                --size;

                                                returntrue;

                                }

                                else

                                {

                                                returnfalse;

                                }

                }

                boolTest(size_tdata)

                {

                                //找数据对应的比特位

                                size_tindex =data>> 5;

                                size_tnum =data% 32;

                                //结果为1则存在，结果为0则不存在

                                returnarr[index] & (1 << num);  

                }

                size_tSize()

                {

                                returnsize;

                }

private:

                vector<size_t> arr;

                size_tsize;

};

voidTestBitMap()

{

                BitMapbt(10);

                bt.Set(9);

                bt.Set(5);

                cout <<"9是否存在？"<< bt.Test(9) << endl;   //1

                cout <<"6是否存在？"<< bt.Test(6) << endl;   //0

                //bt.Reset(9);

                //cout <<"9是否存在？"<< bt.Test(9) << endl;

                bt.Set(7);

                bt.Set(7);

                bt.Set(7);

                cout <<"7是否存在？"<< bt.Test(7) << endl;  //1

                cout <<"size="<< bt.Size() << endl;         //3

                bt.Reset(7);

                bt.Reset(7);

                bt.Reset(7);

                cout <<"7是否存在？"<< bt.Test(7) << endl;  //0

                cout <<"size="<< bt.Size() << endl;         //2

}

intmain()

{

                TestBitMap();

                return0;

}