190. Reverse Bits¶ Time: $O(32) = O(1)$ Space: $O(1)$ C++JavaPython 1 2 3 4 5 6 7 8 9 10 11 12class Solution { public: uint32_t reverseBits(uint32_t n) { uint32_t ans = 0; for (int i = 0; i < 32; ++i) if (n >> i & 1) ans |= 1 << 31 - i; return ans; } }; 1 2 3 4 5 6 7 8 9 10 11 12public class Solution { // You need treat n as an unsigned value public int reverseBits(int n) { int ans = 0; for (int i = 0; i < 32; ++i) if ((n >> i & 1) == 1) ans |= 1 << 31 - i; return ans; } } 1 2 3 4 5 6 7 8 9class Solution: def reverseBits(self, n: int) -> int: ans = 0 for i in range(32): if n >> i & 1: ans |= 1 << 31 - i return ans