[LintCode] Rotate Bits

Bit Rotation -—— A rotation (or circular shift) is an operation similar to shift except that the bits that fall off at one end are put back to the other end.

In left rotation, the bits that fall off at left end are put back at right end.

Let n is stored using 8 bits. Left rotation of n = 　11100101 by 3 makes n = 00101111 (Left shifted by 3 and first 3 bits are put back in last ). If n is stored using 16 bits or 32 bits then left rotation of n (000…11100101) becomes 00..0011100101000.

In this problem, you can assume that n was stored in 32 Bits

Example

Given n = 123, d = 4
return 183

 1 public class Solution {
 2     /*
 3      * @param : a number
 4      * @param : digit needed to be rorated
 5      * @return: a number
 6      */
 7     public int leftRotate(int n, int d) {
 8         if(d <= 0) {
 9             return n;
10         }
11         if(d > 31) {
12             d = d % 32;
13         } 
14         int[] bits = new int[32];
15         for(int i = 0; i < 32; i++) {
16             bits[i] = ((n >>> (31 - i)) & 1);
17         }
18         leftRotateArray(bits, d);
19         int result = 0;
20         for(int i = 0; i < 32; i++) {
21             result |= (bits[31 - i] << i);
22         }
23         return result;
24     }
25     private static void reverse(int[] bits, int startIdx, int endIdx) {
26         while(startIdx < endIdx) {
27             bits[startIdx] ^= bits[endIdx];
28             bits[endIdx] ^= bits[startIdx];
29             bits[startIdx] ^= bits[endIdx];
30             startIdx++;
31             endIdx--;
32         }
33     }
34     private void leftRotateArray(int[] bits, int d) {
35         reverse(bits, 0, d - 1);
36         reverse(bits, d, bits.length - 1);
37         reverse(bits, 0, bits.length - 1);
38     }
39 }