题目描述

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?

Related problem: Reverse Integer

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

分析

JDK ：int reverse(int i)

    /** * Returns the value obtained by reversing the order of the bits in the * two's complement binary representation of the specified {@code int} * value. * * @param i the value to be reversed * @return the value obtained by reversing order of the bits in the * specified {@code int} value. * @since 1.5 */
    public static int reverse(int i) {
        // HD, Figure 7-1
        i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
        i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
        i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
        i = (i << 24) | ((i & 0xff00) << 8) |
            ((i >>> 8) & 0xff00) | (i >>> 24);
        return i;
    }

思路

把第i位取出，放到结果的第31-i位上。

代码

    public static int reverseBits(int n) {

        int rt = 0;

        for (int i = 0; i < 32; i++) {
            rt |= ((n >> i) & 1) << (31 - i);
        }
        return rt;
    }