BitMap

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/** * 在2.5亿个整数中找出不重复的整数，内存不足以容纳这2.5亿个整数。 * 每个数分配2bit，00表示不存在，01表示出现一次，10表示多次，11无意义 *  * <br>参考网址：http://www.cnblogs.com/dongxi/archive/2012/11/07/2759618.html */public class BitMap {//用char数组存储2-Bitmap,不用考虑大小端内存的问题 ,数组大小自定义private static char[] flags = new char[1000];/** * 检查@idx是否已存在 * @param idx * @return */public static int get_val(int idx) {/* * |   8 bit   | * |00 00 00 00|  //映射3 2 1 0 * |00 00 00 00|  //表示7 6 5 4 * . * . * . * |00 00 00 00| */int i = idx / 4; //一个char 表示4个数int j = idx % 4;//0x3是0011 j的范围为0-3，因此0x3<<(2*j)范围为00000011到11000000int ret = (flags[i] & (0x3 << (2 * j))) >> ( 2 * j );return ret;}public static int set_val(int idx, int val) {int i = idx / 4;int j = idx % 4;int tmp = (flags[i] & ~((0x3 << (2 * j)) & 0xff)) | (((val % 4) << ( 2 * j)) & 0xff);flags[i] = (char) tmp;return 0;}public static int add_one(int idx) {int ret = get_val(idx);if (ret >= 2)return 1;else {set_val(idx, ret + 1);return 0;}}public static void main(String[] args) {System.out.println(2>>3);System.out.println(3>>2);System.out.println(2<<3);System.out.println(3<<2);for (int i = 0; i < 8; i++)System.out.println(i + " --> " + get_val(i));//BitSet set = new BitSet(1000000000);int a[] = { 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 1, 3, 5, 1, 3, 1, 10, 2, 4, 6, 8, 0 };int i = 0;System.out.print("所有数据：");for (i = 0; i < a.length; i++) {System.out.print(a[i] + "  ");add_one(a[i]);}System.out.print("\n只出现一次的数据：");for (i = 0; i < 100; i++) {if (get_val(i) == 1)System.out.print(i + "  ");}}}

#include<stdio.h>#include<memory.h>unsigned char flags[1000];unsigned get_val(int idx){int i = idx / 4;int j = idx % 4;unsigned ret = (flags[i] & (0x3 << (2 * j))) >> (2 * j);return ret;}unsigned set_val(int idx, unsigned int val) {int i = idx / 4;int j = idx % 4;unsigned tmp = (flags[i] &~((0x3<<(2 * j)) & 0xff)) | (((val % 4) << (2 * j)) & 0xff);flags[i] = tmp;return 0;}unsigned add_one(int idx) {if (get_val(idx) >= 2)return 1;else {set_val(idx, get_val(idx) + 1);return 0;}}int a[]= { 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 1, 3, 5, 1, 3, 1, 10, 2, 4, 6, 8, 0 };intmain() {int i;memset(flags, 0, sizeof(flags));printf("原数组为:");          for(i = 0; i < sizeof(a) / sizeof(int); ++i)  {              printf("%d  ", a[i]);              add_one(a[i]);          }          printf("\r\n");                printf("只出现过一次的数:");          for(i = 0; i < 100; ++i)  {              if(get_val(i) == 1)                  printf("%d  ", i);          }          printf("\r\n"); return 0;}

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