数据结构之哈希变形——位图

背景：海量数据处理
笔试题：给四十亿不重复的无符号，整形，没排过序。给一个无符号整数，如何快速判断一个数是否在这四十亿个数中

在这里我们就可以用到我们的哈希思想。
位图：我们只需要表示这个数字存不存在的话，只需要一个比特位即可，所以我们用一个数组来装元素的个数，每个元素都有八个比特位，那么就可以表示八个数字【（用uint64_t的话），因为uint64_t不论在多少位机子下都是八字节】
这样就可以轻松解决问题
下面是代码具体实现
bitmap.h

#pragma once
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stddef.h>
#include <unistd.h>
#include <stdint.h>

typedef uint64_t bitmaptype;
typedef struct bitmap{
    bitmaptype *data;
    uint64_t capacity;
}bitmap;
void bitmapinit(bitmap *bm,uint64_t capacity);
void bitmapdestroy(bitmap *bm);
void bitmapset(bitmap *bm,uint64_t index);
void bitmapunset(bitmap *bm,uint64_t index);
void bitmapfill(bitmap *bm,uint64_t capacity);
void bitmapclear(bitmap *bm,uint64_t capacity);
uint64_t bitmaptest(bitmap *bm,uint64_t index);

bitmap.c

#include <stdio.h>
#include "bitmap.h"

void bitmapinit(bitmap *bm,uint64_t capacity)
{
    if(bm == NULL)
    {
        return;
    }
    bm->capacity = capacity;
    uint64_t size = capacity/(sizeof(bitmaptype)*8)+1;
    bm->data = (bitmaptype *)malloc(sizeof(bitmaptype)*size);
    memset(bm->data,0,sizeof(bitmaptype)*size);
}
//uint64_t getsize(uint64_t capacity)
//{
//    uint64_t ret = capacity/(sizeof(bitmaptype)*8)+1;
//    return ret;
//}
void bitmapdestroy(bitmap *bm)
{
    if(bm == NULL)
    {
        return;
    }
    bm->capacity = 0;
    free(bm->data);
    return;
}
uint64_t bitmaptest(bitmap *bm,uint64_t index)
{
    if(bm == NULL || index >= bm->capacity)
    {
        return;
    }
    uint64_t n,offset;
    n = index/(sizeof(bitmaptype)*8);
    offset = index % (sizeof(bitmaptype)*8);
    u_int64_t ret = bm->data[n] & (0x1ul << offset);
    return ret > 0 ? 1 : 0;


}
//void getoffset(uint64_t index,uint64_t* n,uint64_t *offset)
//{
//    *n = index/(sizeof(bitmaptype)*8);
//    *offset = index %(sizeof(bitmaptype)*8);
//    
//}
void bitmapset(bitmap *bm,uint64_t index)
{
    if(bm == NULL)
    {
        return;
    }
    uint64_t n,offset;
    n = index/(sizeof(bitmaptype)*8);
    offset = index % (sizeof(bitmaptype)*8);
    bm->data[n] |= 0x1ul << offset;
    return;

}
void bitmapunset(bitmap *bm,uint64_t index)
{
    if(bm == NULL)
    {
        return;
    }
    uint64_t n,offset;
    n = index/(sizeof(bitmaptype)*8);
    offset = index % (sizeof(bitmaptype)*8);
    bm->data[n] &=  ~(0x1ul << offset);
    return;

}
void bitmapfill(bitmap *bm,uint64_t capacity)
{
    if(bm == NULL)
    {
        return;
    }
    uint64_t size = capacity/(sizeof(bitmaptype)*8)+1;
    memset(bm->data,0xff,sizeof(bitmaptype)*size);
    return;


}
void bitmapclear(bitmap *bm,uint64_t capacity)
{
    if(bm == NULL)
    {
        return;
    }
    uint64_t size = capacity/(sizeof(bitmaptype)*8)+1;
    memset(bm->data,0x0,sizeof(bitmaptype)*size);
    return;

}
#define HEADER printf("\n============%s=============\n",__FUNCTION__)

void testbitmaptest()
{
    HEADER;
    bitmap bitmap;
    bitmapinit(&bitmap,100);
    int ret = bitmaptest(&bitmap,50);
    printf("excpted ret is 0,actul is %d\n",ret);

}
void testbitmapset()
{
    HEADER;
    bitmap bitmap;
    bitmapinit(&bitmap,100);
    int ret1 = bitmaptest(&bitmap,50);
    printf("excpted ret1 is 0,actul is %d\n",ret1);
    bitmapset(&bitmap,50);
    int ret2 = bitmaptest(&bitmap,50);
    printf("excpted ret2 is 1,actul is %d\n",ret2);

}
void testbitmapunset()
{
    HEADER;
    bitmap bitmap;
    bitmapinit(&bitmap,100);
    int ret1 = bitmaptest(&bitmap,50);
    printf("excpted ret1 is 0,actul is %d\n",ret1);
    bitmapset(&bitmap,50);
    int ret2 = bitmaptest(&bitmap,50);
    printf("excpted ret2 is 1,actul is %d\n",ret2);
    bitmapunset(&bitmap,50);
    int ret3 = bitmaptest(&bitmap,50);
    printf("excpted ret3 is 0,actul is %d\n",ret3);

}
void testbitmapfill()
{
    HEADER;
    bitmap bitmap;
    bitmapinit(&bitmap,100);
    int ret1 = bitmaptest(&bitmap,50);
    printf("excpted ret1 is 0,actul is %d\n",ret1);
    bitmapfill(&bitmap,100);
    int ret2 = bitmaptest(&bitmap,50);
    printf("excpted ret2 is 1,actul is %d\n",ret2);


}
void testbitmapclear()
{
    HEADER;
    bitmap bitmap;
    bitmapinit(&bitmap,100);
    int ret1 = bitmaptest(&bitmap,50);
    printf("excpted ret1 is 0,actul is %d\n",ret1);
    bitmapfill(&bitmap,100);
    int ret2 = bitmaptest(&bitmap,50);
    printf("excpted ret2 is 1,actul is %d\n",ret2);
    bitmapclear(&bitmap,100);
    int ret3 = bitmaptest(&bitmap,50);
    printf("excpted ret3 is 0,actul is %d\n",ret3);


}
int main()
{
    testbitmaptest();
    testbitmapset();
    testbitmapunset();
    testbitmapfill();
    testbitmapclear();
    return;
}

下面就是我们的实验结果

不过位图也有缺点，就是它只能表示这个数字存不存在，而不能表示其他东西。