Reverse bits of a given 32 bits unsigned integer.
Note:
3 and the output represents the signed integer 1073741825.Follow up:
If this function is called many times, how would you optimize it?
Example 1:
Input: n = 00000010100101000001111010011100
Output: 964176192 (00111001011110000010100101000000)
Explanation:The input binary string00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is00111001011110000010100101000000.
Example 2:
Input: n = 11111111111111111111111111111101
Output: 3221225471 (10111111111111111111111111111111)
Explanation:The input binary string11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is10111111111111111111111111111111.
Constraints:
32class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t rev = 0;
for(int i=0;i<32;i++)
{
rev = rev<<1 | (n>>i)&1;
}
return rev;
}
};