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LeetCode 190 Reverse Bits

2015-12-04 20:29 148 查看

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)

[code]    /**
     * 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

[code]    public static int reverseBits(int n) {

        int rt = 0;

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