# Reverse Bits

### 描述​

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

Follow up: If this function is called many times, how would you optimize it?

### 解法 1​

// Reverse Bits// Time Complexity: O(logn), Space Complexity: O(1)public class Solution {    // you need treat n as an unsigned value    public int reverseBits(int n) {        int result = 0;        for (int i = 0; i < 32; ++i) {            if ((n & 1) == 1) {                result = (result << 1) + 1;            } else {                result = result << 1;            }            n = n >> 1;        }        return result;    }}

### 解法 2​

// Reverse Bits// Time Complexity: O(logn), Space Complexity: O(1)public class Solution {    // you need treat n as an unsigned value    public int reverseBits(int n) {        int left = 0;        int right = 31;        while (left < right) {            // swap bit            int x = (n >> left) & 1;            int y = (n >> right) & 1;            if (x != y) {                n ^= (1 << left) | (1 << right);            }            ++left;            --right;        }        return n;    }}