leetcode 190. Reverse Bits | 190. 颠倒二进制位（移位操作，十进制二进制相互转换，打表法）

题目

https://leetcode-cn.com/problems/reverse-bits/

题解

方法1：JDK 自带的 Integer.reverse() 方法源码

    /*** 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 7-1i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;return reverseBytes(i);}

方法2：常规解法

public class Solution {public static void main(String[] args) {Solution solution = new Solution();solution.reverseBits2(-3);}/*** 方法2：打败 100%*/public int reverseBits2(int n) {int res = 0;for (int i = 0; i < 32; i++) {res = (res << 1) + (n >>> i & 1);}return res;}/*** 方法1：通过数组*/public int reverseBits(int n) {//输入是一个十进制整数System.out.print(n);int[] bits = new int[32];int size = 0;for (int i = 31; i >= 0; i--) { // 十进制转二进制bits[size++] = n >>> i & 1;}System.out.println(" 转换为二进制：" + Arrays.toString(bits));int res = 0;for (int i = 31; i >= 0; i--) {//二进制转十进制res = ((res << 1) + (bits[i] & 1));}return res;}
}

方法3：打表

1、打 256 表

class Solution {private static final int[] lookup = new int[]{0, 128, 64, 192, 32, 160, 96, 224, 16, 144, 80, 208, 48, 176, 112, 240, 8, 136, 72, 200, 40, 168, 104, 232, 24, 152, 88, 216, 56, 184, 120, 248, 4, 132, 68, 196, 36, 164, 100, 228, 20, 148, 84, 212, 52, 180, 116, 244, 12, 140, 76, 204, 44, 172, 108, 236, 28, 156, 92, 220, 60, 188, 124, 252, 2, 130, 66, 194, 34, 162, 98, 226, 18, 146, 82, 210, 50, 178, 114, 242, 10, 138, 74, 202, 42, 170, 106, 234, 26, 154, 90, 218, 58, 186, 122, 250, 6, 134, 70, 198, 38, 166, 102, 230, 22, 150, 86, 214, 54, 182, 118, 246, 14, 142, 78, 206, 46, 174, 110, 238, 30, 158, 94, 222, 62, 190, 126, 254, 1, 129, 65, 193, 33, 161, 97, 225, 17, 145, 81, 209, 49, 177, 113, 241, 9, 137, 73, 201, 41, 169, 105, 233, 25, 153, 89, 217, 57, 185, 121, 249, 5, 133, 69, 197, 37, 165, 101, 229, 21, 149, 85, 213, 53, 181, 117, 245, 13, 141, 77, 205, 45, 173, 109, 237, 29, 157, 93, 221, 61, 189, 125, 253, 3, 131, 67, 195, 35, 163, 99, 227, 19, 147, 83, 211, 51, 179, 115, 243, 11, 139, 75, 203, 43, 171, 107, 235, 27, 155, 91, 219, 59, 187, 123, 251, 7, 135, 71, 199, 39, 167, 103, 231, 23, 151, 87, 215, 55, 183, 119, 247, 15, 143, 79, 207, 47, 175, 111, 239, 31, 159, 95, 223, 63, 191, 127, 255};public int reverseBits(int n) {int r = 0;for (int i = 0; i < 4; i++) {r = (r << 8) | lookup[n & 0xff];n >>= 8;}return r;}
}

2、打 65536 表

因为代码长度限制，所以 lookup 只能在创建类时计算得到，无法硬编码。

public class Solution {// you need treat n as an unsigned valueprivate static final int[] lookup = new int[65536];{for (int i = 0; i < 65536; i++) {lookup[i] = Integer.reverse(i) >>> 16;}}public int reverseBits(int n) {int r = 0;for (int i = 0; i < 2; i++) {r = (r << 16) | lookup[n & 0xffff];n >>= 16;}return r;}
}

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leetcode 190. Reverse Bits | 190. 颠倒二进制位（移位操作，十进制二进制相互转换，打表法）

题目

题解

方法1：JDK 自带的 Integer.reverse() 方法源码

方法2：常规解法

方法3：打表

1、打 256 表

2、打 65536 表

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