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Counting number of enabled bits in uint32_t.. Here is a O(lg32) solution. Very smart, like bottom-up parallel tree post-order traversal.
Solution from discussion of LC:
int hammingWeight(uint32_t n)
{ n = ((n >> (1 << 0)) & 0x55555555) + (n & 0x55555555); n = ((n >> (1 << 1)) & 0x33333333) + (n & 0x33333333); n = ((n >> (1 << 2)) & 0x0F0F0F0F) + (n & 0x0F0F0F0F); n = ((n >> (1 << 3)) & 0x00FF00FF) + (n & 0x00FF00FF); return ((n >> (1 << 4)) & 0xFFFF) + (n & 0xFFFF); }
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原文地址：http://www.cnblogs.com/tonix/p/4331405.html
