题目:
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?
提示:
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。
代码:
使用bitset:
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
bitset<> bits(n);
int i = , j = , 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 = ;
for (int i = ; i < ; ++i, n = n >> )
m = (m << ) + (n & );
return m;
}
};