Next Power of 2

转载于 2013-06-09 14:11:06 发布 · 2.5k 阅读

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本文介绍两种高效的方法来找出大于等于给定数的最近的2的幂次方数。第一种方法通过逐步左移的方式找到目标数值；第二种方法采用位操作快速实现。这两种算法对于处理与2的幂有关的问题非常实用。

reference:

http://www.geeksforgeeks.org/next-power-of-2/

Problem Definition:

Write a function that, for a given no n, finds a number p which is greater than or equal to n and is a power of 2.

Solution:

1.Shift result one by one, this is straight-forward

2.Customized and Fast

    2.1. Subtract n by 1
       n = n -1

    2.2. Set all bits after the leftmost set bit.

    /* Below solution works only if integer is 32 bits */
                n = n | (n >> 1);
                n = n | (n >> 2);
                n = n | (n >> 4);
                n = n | (n >> 8);
                n = n | (n >> 16);
    2.3. Return n + 1

Code:

1.Shift result one by one, this is straight-forward
unsigned int nextPowerOf2(unsigned int n)
{
    unsigned int p = 1;
    if (n & !(n & (n - 1)))
        return n;
 
    while (p < n) {
        p <<= 1;
    }
    return p;
}


2.Customized and Fast

/* Finds next power of two for n. If n itself
   is a power of two then returns n*/
 
unsigned int nextPowerOf2(unsigned int n)
{
    n--;
    n |= n >> 1;
    n |= n >> 2;
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
    n++;
    return n;
}