0190. Reverse Bits (E)

Posted 2020-12-17 mapoos

Reverse Bits (E)

题目

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?

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题意

将32位无符号整数的二进制数字逆置，返回构成的新整数。

思路

每次取一位逆序处理。

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代码实现

Java

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

javascript

API

/**
 * @param {number} n - a positive integer
 * @return {number} - a positive integer
 */
var reverseBits = function(n) {
  return Number.parseInt(n.toString(2).padStart(32, ‘0‘).split(‘‘).reverse().slice(0, 32).join(‘‘), 2)
};

位运算

/**
 * @param {number} n - a positive integer
 * @return {number} - a positive integer
 */
var reverseBits = function (n) {
  let ans = 0
  for (let i = 0; i < 32; i++) {
    ans <<= 1
    ans |= n & 1
    n >>= 1
  }
  ans >>>= 0		// js位运算默认使用有符号整数，该步骤可以将其转化为无符号整数
  return ans
}