leetcode+位操作，求反

点击打开链接 class Solution { public: uint32_t reverseBits(uint32_t n) { uint32_t result = 0; int bit = 0; for(int i=0; i<32; i++){ bit = n&1;//一次求出从低位到高位的数字来 n = n>>1; result = 2*result+bit; } return result; } };