把一个32位的二进制进行逆序处理-xwjazjx1314-ChinaUnix博客


Also check http://massivealgorithms.blogspot.com/2014/06/reverse-integer-leetcode-easycpp.html
把一个32位的二进制进行逆序处理-xwjazjx1314-ChinaUnix博客
设想一下,逆序'ab',为'ba'。
逆序abcd,可以先两两交换为cdab,然后一一交换为dcba。
逆序abcdefgh,先四四交换efghabcd,然后两两交换fehgcdab,然后一一交换efghdcaba。
那么可以推广到二进制表示:
第一步:每2位为一组,组内高低位交换
       00 11 00 00 00 11 10 01
  -->00 11 00 00 00 11 01 10
第二步:每4位为一组,组内高低位交换
       0011 0000  0011 0110
  -->1100 0000 1100 1001
第三步:每8位为一组,组内高低位交换
       11000000 11001001
  -->00001100 10011100
第四步:每16位为一组,组内高低位交换
       0000110010011100
  -->1001110000001100
高低位互换时操作大概就是偶数位左移1位,奇数位右移1位
      原 数   00110000 00111001
      奇数位 0_1_0_0_ 0_1_1_0_
      偶数位 _0_1_0_0 _0_1_0_1
其余位数用0填充,然后将奇数位右移一位,偶数位左移一位,此时将这两个数据做按位与运算,即可以达到奇偶位上数据交换的效果了:
    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;
    }
def reverseBinary(a):
    a = ((a & 0xAAAA) >> 1) | ((a & 0x5555) << 1)
    a = ((a & 0xCCCC) >> 2) | ((a & 0x3333) << 2) 
    a = ((a & 0xF0F0) >> 4) | ((a & 0x0F0F) << 4) 
    a = ((a & 0xFF00) >> 8) | ((a & 0x00FF) << 8)
    return a
  1. int main() {
  2.     int n;
  3.     while (scanf("%d", &n) != EOF) {
  4.         n = (& 0x55555555) << 1 | (& 0xAAAAAAAA) >> 1;
  5.         n = (& 0x33333333) << 2 | (& 0xCCCCCCCC) >> 2;
  6.         n = (& 0x0F0F0F0F) << 4 | (& 0xF0F0F0F0) >> 4;
  7.         n = (& 0x00FF00FF) << 8 | (& 0xFF00FF00) >> 8;
  8.         n = (& 0x0000FFFF) << 16 | (& 0xFFFF0000) >> 16;
  9.         printf ("%d\n", n);
  10.     }
  11.     return 0;
  12. }
  13. 其实,这种方法很好理解,上面的意思就是通过移位,先两两交换。看下图就明白其中的原理了。
  14. 假如一个8位的二进制数,就是如图所示:11 10 01 00 先1跟1交换,1跟0交换,0跟1交换 0跟0交换。
  15. 第二步就是11 与01交换。。。最后,得出所需要的结果。 16位及32位类似。
                
     
这种方法的好处就是:直观明了,一步一步的往下执行。

最直接的方法

最简单的做法,原数不断右移取出最低位,赋给新数的最低位后新数再不断左移。
    public int reverseBits(int n) {
        int res = 0;
        for(int i = 0; i < 32; i++, n >>= 1){
            res = res << 1 | (n & 1);
        }
        return res;
    }
最容易想到的方法就是依次交换两端的数据,从右向左遍历数字,当i位遇到1时,将逆序数字对应的(17-i)位设为1。
def reverseBinary(num):
    print bin(num)
    new=0
    tmp=(1<<15)
    for i in xrange(16):
        if num&1:
            new|=tmp
        tmp>>=1
        num>>=1
    return new

Convert to char array ==> not good
void bitreverse(char *s, int x)
{
    int i = 0;
    for(; i < INTBIT; x >>= 1)
    {
        s[i++] = x & 1 ? '1' : '0';
    }
    s[i] = '\0';
}

  1. int reverse( int n )
  2. {
  3.     unsigned int tmp;
  4.     unsigned int maskl=0x01;
  5.     unsigned int maskh=0x80000000;
  6.     unsigned int rst=0;
  7.     int i,j;
  8.     for(i=0;i<15;i++)
  9.     {
  10.           tmp=n&maskl;
  11.           maskl=maskl<<1;
  12.           for(j=i+1;j<=31-i;j++)
  13.                 tmp=tmp<<1;
  14.           rst=tmp | rst;
  15.     }
  16.     for(i=0;i<15;i++)
  17.     {
  18.           tmp=n&maskh;
  19.           maskl=maskh>>1;
  20.           for(j=i+1;j<=31-i;j++)
  21.                 tmp=tmp>>1;
  22.           rst=tmp | rst;
  23.     }
  24.     return rst;
  25. }
这种方法也是类似。先取出高位或者低位,再按位或。
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