190. Reverse Bits(Solution || Leetcode easy || Java)

2 min readDec 20, 2022

Reverse bits of a given 32 bits unsigned integer.


  • Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.
  • In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

Example 1:

Input: n = 00000010100101000001111010011100
Output: 964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input: n = 11111111111111111111111111111101
Output: 3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.


  • The input must be a binary string of length 32


public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
//Create a integer to store answer
int ans = 0;

//run the loop 32 times as it is given in constraints that
//the int is 32 bit long
for(int i = 0; i < 32; i++){

//append the last bit of integer to reverse it
ans |= (n & 1);

//this condition is necessary anotherwise the loop will
// add a extra zero
if(i != 31)

//left shift ans to store next bit
ans = ans << 1;

//as the last bit is stored in ans remove that bit
n = n >> 1;

return ans;

Runtime 1 ms



Memory 42.6 MB



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