Reverse Bits

描述

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

Follow up: If this function is called many times, how would you optimize it?

分析

最简单直接的做法，从右向左把一位位取出来，添加到新生成的整数的最低位即可。

第二个简单的方法，左右不断交换位，直到相遇。

解法1

  1. // Reverse Bits
  2. // Time Complexity: O(logn), Space Complexity: O(1)
  3. public class Solution {
  4.     // you need treat n as an unsigned value
  5.     public int reverseBits(int n) {
  6.         int result = 0;
  7.         for (int i = 0; i < 32; ++i) {
  8.             if ((n & 1) == 1) {
  9.                 result = (result << 1) + 1;
  10.             } else {
  11.                 result = result << 1;
  12.             }
  13.             n = n >> 1;
  14.         }
  15.         return result;
  16.     }
  17. }

解法2

  1. // Reverse Bits
  2. // Time Complexity: O(logn), Space Complexity: O(1)
  3. public class Solution {
  4.     // you need treat n as an unsigned value
  5.     public int reverseBits(int n) {
  6.         int left = 0;
  7.         int right = 31;
  8.         while (left < right) {
  9.             // swap bit
  10.             int x = (n >> left) & 1;
  11.             int y = (n >> right) & 1;
  13.             if (x != y) {
  14.                 n ^= (1 << left) | (1 << right);
  15.             }
  16.             ++left;
  17.             --right;
  18.         }
  19.         return n;
  20.     }
  21. }

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/bitwise-operations/reverse-bits.html