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?
这道题比较简单，考的就是按位操作，可以借助于STL的bitset，也可以直接通过位运算完成题目的要求。
使用bitset:
class Solution { public: uint32_t reverseBits(uint32_t n) { bitset<32> bits(n); int i = 0, j = 31, tmp; for (; i < j; ++i, --j) { tmp = bits[i]; bits[i] = bits[j]; bits[j] = tmp; } unsigned long l = bits.to_ulong(); return (uint32_t)l; } };
直接按位操作：
class Solution { public: uint32_t reverseBits(uint32_t n) { uint32_t m = 0; for (int i = 0; i < 32 ; ++i, n = n >> 1) m = (m << 1) + (n & 1); return m; } };
原文：http://www.cnblogs.com/jdneo/p/4797197.html