zoukankan      html  css  js  c++  java
  • 190. Reverse Bits

    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?

    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;
        }
    }
  • 相关阅读:
    fedora/centos7防火墙FirewallD详解
    python for dl
    神经网络画图工具
    卷积神经网络的复杂度分析
    如何理解深度学习中的Transposed Convolution?
    吴恩达课程及视频笔记汇总
    从LeNet-5到DenseNet
    WPS for Linux
    caffe:fine-tuning
    python下图像读取方式以及效率对比
  • 原文地址:https://www.cnblogs.com/fatttcat/p/11404502.html
Copyright © 2011-2022 走看看