（Java）leetcode-190 Reverse Bits

题目

【按位反转】
Reverse bits of a given 32 bits unsigned integer.

Example 1:

Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
Example 2:

Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.

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 signed integer type and should not affect your implementation, as the internal binary representation of the integer 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.

Follow up:

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

思路

涉及到二进制，想到按位操作，新建一个变量用来存新的值，原值一边右移（注意必须要用无符号右移，不然若是负数，高位会产生多余的“1”），一边按位与（&)，这样就能依次将原值二进制的每一位取出来，然后新值一边左移一边+取出来的值，达到反转的效果。最后补上0即可。
时间复杂度：O(n)

代码

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int j = 0,num = 0;
        
        while(n!=0){
        	j  <<= 1;
            num++;
            j += (n & 1);
        	n >>>= 1;
            
        }        
        j <<= (32-num);
        return j;
    }
}

提交结果

Runtime: 1 ms, faster than 100.00% of Java online submissions for Reverse Bits.
Memory Usage: 27.9 MB, less than 91.88% of Java online submissions for Reverse Bits.