ADT-位图

利用位图数据结构实现排序，利用每一位的下标作为索引，每位的值作为属性值，可以表示存在或不存在，适合存储稠密的数据，排序遍历的范围会是索引的最大值

最后发现耗时比python中list自带的sort多，

 # -*- coding:utf-8 -*-
 import time
 class BitMap:                                   #位图
         def __init__(self,size):
                 self.size = size
                 self.byteSize = (size-1)/32+2
                 self.array = [0 for i in range(self.byteSize)]

         def set(self,bit):
                 index=bit >> 5
                 bit = 31 - bit & 31
                 p = 1 << bit
                 self.array[index] = self.array[index] | p

         def cls(self,bit):
                 index=bit >> 5
                 bit = 31 - bit & 31
                 p = ~ (1 << bit)
                 self.array[index] = array[index] & p

         def get(self,bit):
                 index=bit >> 5
                 bit = 31 - bit & 31
                 p = 1 << bit
                 n=self.array[index] & p
                 n = n >> bit
                 return n

 Num = 10000000
 a = BitMap(Num)
 fr = open("out.txt","r")
 out = open('new1.txt',"w")
 Str = fr.read()
 List = str.split(Str,' ')
 start = time.clock()

 for i in range(len(List)-1):
         List[i]=int(List[i])
         a.set(List[i])
 for i in range(Num+1):
     if a.get(i):
         out.write(str(i)+' ')
 '''
 for i in range(len(List)-1):
         List[i]=int(List[i])
 List.sort()
 for i in List:
         out.write(str(i))
         out.write(' ')
 '''
 end = time.clock()
 print '位图排序：',end - start
 fr.close()
 out.close()

精简版ADT实现

 #define BITSPERWORF 32
 #define SHIFT 5
 #define MASK 0x1f
 #define N 1e7
 +N/BITSPERWORF]

 void set(int i){
     a[>>SHIFT] |= (<<(i&MASK));
 }
 void clr(int i){
     a[>>SHIFT] &= ~(<<(i&MASK));
 }
 void test(int i){
     >>SHIFT] & (<<(i&MASK));
 }