Reverse bits of a given 32 bits unsigned integer.

Note:

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.

<解題>

1. 利用n&1，將 n 的最後一位（最低位）取出，可以使用 n & 1 進行位元運算，即取 n 的最後一位的數值（0 或 1）
2. 使用位元 OR 運算將取得的最後一位數值加到 result 上（result | (n & 1)），這相當於將 n 的最後一位數值放入 result 的最低位
3. 將 n 向右移位（n = n » 1），相當於將 n 的二進制表示向右移動一位，並忽略最低位的數值。這樣在下一次迴圈中，就可以繼續取得 n 的下一位。

public class Solution {
  public int reverseBits(int n) {
    int ans=0;
    for(int i=0;i<32;i++){
    	ans= ( ans << 1 ) | ( n & 1 );         
    	n = n >> 1;                  
    }
    return ans;
}
}

Time: O(1)

Space: O(1)