Brooke

一、題目

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.

 

Constraints:

• The input must be a binary string of length 32

 

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

 

二、程式作法

 /*寫法二*/
public class Solution
{
    public uint reverseBits(uint n)
    {
        uint result = 0;
        uint[] a = new uint[32];
        int counter = 0;

        while (n > 0)
        {
            a[counter] = n % 2;
            n /= 2;
            counter++;
        }

        counter = 0;
        while (counter < a.Length / 2)
        {
            uint temp = a[counter];
            a[counter] = a[a.Length - 1 - counter];
            a[a.Length - 1 - counter] = temp;
            counter++;
        }

        counter = 0;
        while (counter < a.Length)
        {
            result += (uint)Math.Pow(2, counter) * a[counter];
            counter++;
        }

        return result;
    }
}

/*寫法一*/
/*
public class Solution
{
    public uint reverseBits(uint n)
    {
        string s = Convert.ToString(n, 2);
        char[] c = s.ToCharArray();//index 越大，權重越低位，反之，index 在起始為零時，為最大權重
        char[] c32 = new char[32];

        for (int i = 0; i < c32.Length; i++)
        {
            if (i + c.Length < c32.Length)
                c32[i] = '0';
            else
                c32[i] = c[i - (c32.Length - c.Length)];
        }

        for (int i = 0; i < c32.Length / 2; i++)
        {
            char temp = c32[i];
            c32[i] = c32[c32.Length - 1 - i];
            c32[c32.Length - 1 - i] = temp;
        }

        string s32 = new string(c32);

        return Convert.ToUInt32(s32, 2);
    }
}
*/

 

三、思路筆記

我想法為將該無號整數先轉成陣列，然後再做元素調換，最後轉回無號整數回傳答案。