bit-wise reverse

lcf 2014-05-27 05:08:39

想起多年前听朋友叙述其面试题目，说怎么做一个bit-wise reverse （10101010 -> 01010101）在O(log(n))内解决 n = number of bits。没有任何其他限制。当时还记得的算法现在不记得了，也想不出来了（老了吧。。。），谁给提个醒儿呗

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lcf 2014-05-28

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多谢各位提醒！散分

俺是小王子 2014-05-27

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引用 6 楼 vnvlyp 的回复:

[quote=引用 5 楼 benma378 的回复:] 我只是感觉有点抽象，通过那样的位运算就倒转过来了，实在没有看明白，能不能说一下数学原理

以i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;为例，你懂了这个就应该知道其他几行在干嘛先看 | 或运算左边的(i & 0x55555555) << 1，首先0x55555555转换为二进制就是01010101010101010101010101010101，用 i 和这个数做&与运算，就是保留 i 所有奇数位，而把所有偶数位变为0，然后<<1是左移一位，左移后所有奇数位就左移到了偶数位上，再看右边的(i >>> 1) & 0x55555555，首先>>>1是无符号右移，右移以后 i 所有偶数位就都移到了奇数位，然后和0x55555555就是刚才那个数进行&运算，那么同样保留所有奇数位，偶数位变0，注意现在的奇数位实际上是 i 的偶数位。最后进行 | 或运算，把移动后的奇数位偶数位合并起来，所以最终效果就是原来奇数位的移动到偶数位上去了，原来偶数位的移动到奇数位去了。 0x33333333等是一样的原理，它的二进制为00110011001100110011001100110011，可以看到它的作用是每两位之间的移动， 0x0F0F0F0F则是00001111000011110000111100001111，每四位移动[/quote] 谢谢～

vnvlyp 2014-05-27

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引用 5 楼 benma378 的回复:

我只是感觉有点抽象，通过那样的位运算就倒转过来了，实在没有看明白，能不能说一下数学原理

俺是小王子 2014-05-27

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引用 3 楼 fatg1988 的回复:

[quote=引用 2 楼 benma378 的回复:] [quote=引用 1 楼 fatg1988 的回复:]

/**
     * Returns the value obtained by reversing the order of the bits in the
     * two's complement binary representation of the specified <tt>int</tt>
     * value.
     *
     * @return the value obtained by reversing order of the bits in the
     *     specified <tt>int</tt> value.
     * @since 1.5
     */
    public static int reverse(int i) {
        // HD, Figure 7-1
	i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
	i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
	i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
	i = (i << 24) | ((i & 0xff00) << 8) |
	    ((i >>> 8) & 0xff00) | (i >>> 24);
	return i;
    }

jdk里面的分治法 01010101 00110011 00001111 这就是 3 5 F 能看明白吗？

我没看明白。。。。

能否详细解释一下。。。。谢谢[/quote] 比如0101这个 i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555; 这是交换2位的拆解成 (i & 0x5) << 1 | (i >>> 1) & 0x5; 这样你看清楚一点完成的是2位的交换 [/quote] 效果上是这样的，但是我还是没有想明白这原理到底是什么？感觉有点抽象？

引用 4 楼 vnvlyp 的回复:

1L正解，其实就是二分法举个例子，要倒转11010101 只需要将其一分为二即1101和0101，分别倒转这两部分，然后交换位置就可以了 1101 倒转变成 1011 0101 倒转变成 1010 再交换位置得结果10101011 至于怎么倒转那两部分，同样的二分方法，最后到你只需要交换两位的时候就相当简单了因为是二分法，所以复杂度Θ(log(n))

求解释，表示我还没有看出二分法。。。。我只是感觉有点抽象，通过那样的位运算就倒转过来了，实在没有看明白，能不能说一下数学原理谢谢！或者有什么样的资料可以看看？谢谢

vnvlyp 2014-05-27

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1L正解，其实就是二分法举个例子，要倒转11010101 只需要将其一分为二即1101和0101，分别倒转这两部分，然后交换位置就可以了 1101 倒转变成 1011 0101 倒转变成 1010 再交换位置得结果10101011 至于怎么倒转那两部分，同样的二分方法，最后到你只需要交换两位的时候就相当简单了因为是二分法，所以复杂度Θ(log(n))

七神之光 2014-05-27

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引用 2 楼 benma378 的回复:

[quote=引用 1 楼 fatg1988 的回复:]

/**
     * Returns the value obtained by reversing the order of the bits in the
     * two's complement binary representation of the specified <tt>int</tt>
     * value.
     *
     * @return the value obtained by reversing order of the bits in the
     *     specified <tt>int</tt> value.
     * @since 1.5
     */
    public static int reverse(int i) {
        // HD, Figure 7-1
	i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
	i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
	i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
	i = (i << 24) | ((i & 0xff00) << 8) |
	    ((i >>> 8) & 0xff00) | (i >>> 24);
	return i;
    }

jdk里面的分治法 01010101 00110011 00001111 这就是 3 5 F 能看明白吗？

我没看明白。。。。

能否详细解释一下。。。。谢谢[/quote] 比如0101这个 i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555; 这是交换2位的拆解成 (i & 0x5) << 1 | (i >>> 1) & 0x5; 这样你看清楚一点完成的是2位的交换

俺是小王子 2014-05-27

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引用 1 楼 fatg1988 的回复:

/**
     * Returns the value obtained by reversing the order of the bits in the
     * two's complement binary representation of the specified <tt>int</tt>
     * value.
     *
     * @return the value obtained by reversing order of the bits in the
     *     specified <tt>int</tt> value.
     * @since 1.5
     */
    public static int reverse(int i) {
        // HD, Figure 7-1
	i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
	i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
	i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
	i = (i << 24) | ((i & 0xff00) << 8) |
	    ((i >>> 8) & 0xff00) | (i >>> 24);
	return i;
    }

jdk里面的分治法 01010101 00110011 00001111 这就是 3 5 F 能看明白吗？

我没看明白。。。。

能否详细解释一下。。。。谢谢

七神之光 2014-05-27

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/**
     * Returns the value obtained by reversing the order of the bits in the
     * two's complement binary representation of the specified <tt>int</tt>
     * value.
     *
     * @return the value obtained by reversing order of the bits in the
     *     specified <tt>int</tt> value.
     * @since 1.5
     */
    public static int reverse(int i) {
        // HD, Figure 7-1
	i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
	i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
	i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
	i = (i << 24) | ((i & 0xff00) << 8) |
	    ((i >>> 8) & 0xff00) | (i >>> 24);
	return i;
    }

jdk里面的分治法 01010101 00110011 00001111 这就是 3 5 F 能看明白吗？

This is a book for the computationalist, whether a working programmer or anyone interested in methods of computation. The focus is on material that does not usually appear in textbooks on algorithms.

Table of Contents Preface 1 Chapter 1: Shell Something Out 7 Introduction 7 Printing in the terminal 9 Playing with variables and environment variables 12 Doing math calculations with the shell 17 Playing with file descriptors and redirection 19 Arrays and associative arrays 25 Visiting aliases 27 Grabbing information about terminal 29 Getting, setting dates, and delays 30 Debugging the script 33 Functions and arguments 35 Reading the output of a sequence of commands 38 Reading "n" characters without pressing Return 40 Field separators and iterators 41 Comparisons and tests 44 Chapter 2: Have a Good Command 49 Introduction 50 Concatenating with cat 50 Recording and playback of terminal sessions 53 Finding files and file listing 55 Playing with xargs 63 Translating with tr 69 Checksum and verification 72 Sorting, unique and duplicates 75 Temporary file naming and random numbers 80 Splitting files and data 81 ii Table of Contents Slicing file names based on extension 84 Renaming and moving files in bulk 86 Spell checking and dictionary manipulation 89 Automating interactive input 90 Chapter 3: File In, File Out 95 Introduction 96 Generating files of any size 96 Intersection and set difference (A-B) on text files 97 Finding and deleting duplicate files 100 Making directories for a long path 103 File permissions, ownership, and sticky bit 104 Making files immutable 109 Generating blank files in bulk 110 Finding a symbolic link and its target 111 Enumerating file type statistics 113 Loopback files and mounting 115 Creating ISO files, Hybrid ISO 117 Finding difference between files, patching 120 head and tail – printing the last or first 10 lines 122 Listing only directories – alternative methods 125 Fast command-line navigation using pushd and popd 126 Counting number of lines, words, and characters in a file 128 Printing directory tree 129 Chapter 4: Texting and Driving 131 Introduction 132 Basic regular expression primer 132 Searching and mining "text" inside a file with grep 136 Column-wise cutting of a file with cut 142 Frequency of words used in a given file 146 Basic sed primer 147 Basic awk primer 150 Replacing strings from a text or file 156 Compressing or decompressing JavaScript 158 Iterating through lines, words, and characters in a file 161 Merging multiple files as columns 162 Printing the nth word or column in a file or line 163 Printing text between line numbers or patterns 164 Checking palindrome strings with a script 165 Printing lines in the reverse order 169 Parsing e-mail addresses and URLs from text 171 iii Table of Contents Printing n lines before or after a pattern in a file 172 Removing a sentence in a file containing a word 174 Implementing head, tail, and tac with awk 175 Text slicing and parameter operations 177 Chapter 5: Tangled Web? Not At All! 179 Introduction 180 Downloading from a web page 180 Downloading a web page as formatted plain text 183 A primer on cURL 183 Accessing Gmail from the command line 188 Parsing data from a website 189 Image crawler and downloader 191 Web photo album generator 193 Twitter command-line client 195 define utility with Web backend 197 Finding broken links in a website 199 Tracking changes to a website 200 Posting to a web page and reading response 203 Chapter 6: The Backup Plan 205 Introduction 205 Archiving with tar 206 Archiving with cpio 211 Compressing with gunzip (gzip) 212 Compressing with bunzip (bzip) 215 Compressing with lzma 217 Archiving and compressing with zip 219 squashfs – the heavy compression filesystem 220 Cryptographic tools and hashes 222 Backup snapshots with rsync 224 Version control based backup with Git 227 Cloning hard drive and disks with dd 230 Chapter 7: The Old-boy Network 233 Introduction 233 Basic networking primer 234 Let's ping! 241 Listing all the machines alive on a network 243 Transferring files 247 Setting up an Ethernet and wireless LAN with script 250 Password-less auto-login with SSH 253 Running commands on remote host with SSH 255

public static int reverse(int i) { i = (i & 0x55555555) >> 1) & 0x55555555; i = (i & 0x33333333) >> 2) & 0x33333333; i = (i & 0x0f0f0f0f) >> 4) & 0x0f0f0f0f; i = (i << 24) | ((i & 0xff00) << 8

参考: http://www.codeproject.com/Articles/28681/Bitwise-Operation-Explained http://blog.csdn.net/haoni123321/article/details/7415498 //=============Let`s begin================//bitwise operation trick交换两

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