[刷题防痴呆] 0190 - 颠倒二进制位 (Reverse B

题目地址

https://leetcode.com/problems/reverse-bits/

题目描述

190. Reverse Bits

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.

思路

• res左移一位, 留出可以放新一位的位置.
• 然后res加上n & 1, 即当前最低位的bit.
• n右移一位, 去掉已经放到res里的最低位.

关键点

代码

• 语言支持：Java

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int res = 0;
        for (int i = 0; i < 32; i++) {
            res <<= 1;
            res += n & 1;
            n >>= 1;
        }
        return res;
    }
}