Bit manipulation

137. Single Number II

An elegant solution from [1]

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        low_bits = high_bits = 0
        for num in nums:
            low_bits = ~high_bits & (low_bits ^ num)
            high_bits = ~low_bits & (high_bits ^ num)
        return low_bits

Another nice explanation about this nice solution in Chinese can be found in [2]

190. Reverse Bits

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

Reference

[1]https://medium.com/@lenchen/leetcode-137-single-number-ii-31af98b0f462

[2]https://blog.csdn.net/yutianzuijin/article/details/50597413?utm_medium=distribute.pc_relevant.none-task-blog-baidujs-1