190. Reverse Bits

problem

solution

利用 >>i 取得第i位元

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t ret = 0;
        for(int i=0;i<32;++i){
            ret<<=1;
            ret += (n>>i)&1;
        }
        return ret;
    }
};

analysis

• time complexity O(1)
• space complexity O(1)