## 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;
}
}