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Reverse Bits
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Reverse bits of a given 32 bits unsigned integer.
For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).
Follow up:
If this function is called many times, how would you optimize it?
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将一个数的二进制反过来，求反过来的数。
期，将该数的最低位，作为最高位即可（2^31）,然后依次求解。
例： 对于 0000..........10 ans = 0 ans += 0*(2^31) ans += 1*(2^30) ... ...
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t i = 1 ,ans = 0,m = 31; // 要用uint32_t 不能用int
while (m --) { // 将 i = 2^31
i <<= 1;
}
m = 32;
while (n) { // m -- 用m-- 和n 都可以
ans += (n&1)*i;
i >>= 1;
n >>= 1;
}
return ans;
}
};
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