[算法]从十亿个数中找出不重复的数字

 

 

 

 1 import java.util.*;
 2 
 3 public class SelectOnlys {
 4     long S1 = 2 ^ 16;
 5 
 6 
 7     int whichSeg(long s) {
 8         long L = S1;
 9         int flag = 0;
10         while (s >= L) {
11             s %= L;
12             flag++;
13         }
14         return flag;
15     }
16 
17     Collection<Integer> run(long[] a) {
18         int[] count = new int[2 ^ 16];
19         Collection<Integer> res = new ArrayList<Integer>();
20         for (int i = 0; i < a.length; i++) {
21             int seg = whichSeg(a[i]);
22             int rem = (int) (a[i] % S1);
23 
24             if ((count[rem] & (1 << (2 * seg))) != 1) {
25                 count[rem] |= 1 << (2 * seg);
26             } else if ((count[rem] & (1 << (1 + 2 * seg))) != 1) {
27                 count[rem] |= 1 << (1 + 2 * seg);
28             }
29 
30         }
31 
32         for (int i = 0; i < a.length; i++) {
33             int[] b = {1, 3, 5, 7};
34             for (int one : b) {
35                 if ((count[i] & (1 << one)) != 1 && (count[i] & (1 << (one - 1))) == 1) {
36                     res.add((int) (long) (one * S1 + a[i]));
37                 }
38             }
39 
40         }
41         return res;
42 
43     }
44 
45 }

 

posted @ 2017-09-17 22:02  wilderness  阅读(1199)  评论(0编辑  收藏  举报