加入收藏 | 设为首页 | 会员中心 | 我要投稿 李大同 (https://www.lidatong.com.cn/)- 科技、建站、经验、云计算、5G、大数据,站长网!
当前位置: 首页 > 编程开发 > Java > 正文

190. Reverse Bits - Easy

发布时间:2020-12-15 07:57:41 所属栏目:Java 来源:网络整理
导读:Reverse bits of a given 32 bits unsigned integer. ? Example 1: Input: 00000010100101000001111010011100Output: 00111001011110000010100101000000Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned int

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: 22222222222222222222222222222201
Output: 10222222222222222222222222222222
Explanation: The input binary string 22222222222222222222222222222201 represents the unsigned integer 4294967293,so return 3221225471 which its binary representation is 10102222210010110010011101101001.

?

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?

?

e.g.
n = b7 1 b5 b4 b3 b2 0 b0
0 0 0 0 0 0 1 0 1 << i (i = 1)
0 1 0 0 0 0 0 0 1 << j (j = 6)
xor 0 1 0 0 0 0 1 0 bit_mask = ( (1 << i) | (1 << j) )
b7 0 b5 b4 b3 b2 1 b0 n ^ bit_mask
b7 ~b6 b5 b4 b3 b2 ~b1 b0

time = O(1),space = O(1)

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int i = 0,j = 31;
        while(i < j) {
            n = reverseBitPair(n,i++,j--);
        }
        return n;
    }
    
    private int reverseBitPair(int n,int i,int j) {
        int leftBit = (n >> i) & 1;
        int rightBit = (n >> j) & 1;
        if(leftBit != rightBit) {
            n ^= ((1 << i) | (1 << j));
        }
        return n;
    }
}

(编辑:李大同)

【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容!

    推荐文章
      热点阅读