我对位向量如何工作感到困惑（对位向量不太熟悉）。这是给出的代码。有人可以帮我完成这个吗？
public static boolean isUniqueChars(String str) {
int checker = 0;
for (int i = 0; i < str.length(); ++i) {
int val = str.charAt(i) - 'a';
if ((checker & (1 << val)) > 0) return false;
checker |= (1 << val);
}
return true;
}
特别是在checker做什么？
我有一个偷偷摸摸的怀疑你从我正在阅读的同一本书中得到这个代码......这里的代码本身并不像运算符那么神秘 - |=，&，和 << 通常不使用我们外行 - 作者没有费心花额外的时间来解释这个过程，也没有解释这里涉及的实际机制是什么。一开始我对这个线程上的先前答案感到满意，但只是在抽象层面上。我回到它是因为我觉得需要一个更具体的解释——缺少一个总是让我感到不安。
此运算符 << 是一个左位移位器，它采用该数字或操作数的二进制表示并将其移动到由右侧的操作数或数字指定的许多位置，例如仅在二进制中的十进制数。我们乘以 2 为底——当我们向上移动时，很多地方不是以 10 为底——所以右边的数字是指数，左边的数字是 2 的基数倍数。
此运算符 |=（称为按位或赋值）将左侧的操作数与右侧的操作数进行或运算并将结果分配给左侧的操作数（x |= y 等效于 x = x | y）。同样，运算符（'&'）将'和'左侧的运算符与右侧的运算符。这也有一个按位与赋值（x &= y 等价于 x = x & y）。
所以我们这里有一个哈希表，每次检查器获取或'd ( checker |= (1 << val)) 时，它都存储在一个 32 位二进制数中，其中一个字母的指定二进制值将其对应的位设置为真。字符的值是与检查器 ( checker & (1 << val)) > 0) - 如果它大于 0，我们知道我们有一个欺骗 - 因为两个相同的位设置为 true 并且一起将返回 true 或 '1''。
有 26 个二进制位置，每个位置对应一个小写字母——作者确实说过假设字符串只包含小写字母——这是因为我们只剩下 6 个（32 位整数）位置可供使用——而且比我们发生碰撞
00000000000000000000000000000001 a 2^0
00000000000000000000000000000010 b 2^1
00000000000000000000000000000100 c 2^2
00000000000000000000000000001000 d 2^3
00000000000000000000000000010000 e 2^4
00000000000000000000000000100000 f 2^5
00000000000000000000000001000000 g 2^6
00000000000000000000000010000000 h 2^7
00000000000000000000000100000000 i 2^8
00000000000000000000001000000000 j 2^9
00000000000000000000010000000000 k 2^10
00000000000000000000100000000000 l 2^11
00000000000000000001000000000000 m 2^12
00000000000000000010000000000000 n 2^13
00000000000000000100000000000000 o 2^14
00000000000000001000000000000000 p 2^15
00000000000000010000000000000000 q 2^16
00000000000000100000000000000000 r 2^17
00000000000001000000000000000000 s 2^18
00000000000010000000000000000000 t 2^19
00000000000100000000000000000000 u 2^20
00000000001000000000000000000000 v 2^21
00000000010000000000000000000000 w 2^22
00000000100000000000000000000000 x 2^23
00000001000000000000000000000000 y 2^24
00000010000000000000000000000000 z 2^25
因此，对于输入字符串 'azya'，随着我们一步一步地移动
字符串'a'
a =00000000000000000000000000000001
checker=00000000000000000000000000000000
checker='a' or checker;
// checker now becomes = 00000000000000000000000000000001
checker=00000000000000000000000000000001
a and checker=0 no dupes condition
字符串'az'
checker=00000000000000000000000000000001
z =00000010000000000000000000000000
z and checker=0 no dupes
checker=z or checker;
// checker now becomes 00000010000000000000000000000001
字符串 'azy'
checker= 00000010000000000000000000000001
y = 00000001000000000000000000000000
checker and y=0 no dupes condition
checker= checker or y;
// checker now becomes = 00000011000000000000000000000001
字符串'azya'
checker= 00000011000000000000000000000001
a = 00000000000000000000000000000001
a and checker=1 we have a dupe
现在，它声明了一个重复
int checker在这里用作比特的存储。整数值中的每个位都可以视为一个标志，因此最终int是一个位数组（标志）。代码中的每个位都说明是否在字符串中找到了具有位索引的字符。出于同样的原因，您可以使用位向量而不是int. 它们之间有两个区别：
尺寸。int具有固定大小，通常为 4 个字节，这意味着 8*4=32 位（标志）。位向量通常可以有不同的大小，或者您应该在构造函数中指定大小。
应用程序接口。使用位向量，您将更容易阅读代码，可能是这样的：
vector.SetFlag(4, true); // set flag at index 4 as true
因为int您将拥有较低级别的位逻辑代码：
checker |= (1 << 5); // set flag at index 5 to true
也可能int会快一点，因为位操作是非常低级的，可以由 CPU 按原样执行。BitVector 允许编写更少的神秘代码，而且它可以存储更多的标志。
供将来参考：位向量也称为 bitSet 或 bitArray。以下是针对不同语言/平台的此数据结构的一些链接：
我认为所有这些答案确实解释了它是如何工作的，但是我想通过重命名一些变量，添加一些其他变量并添加评论来提供我如何更好地看待它的意见：
public static boolean isUniqueChars(String str) {
/*
checker is the bit array, it will have a 1 on the character index that
has appeared before and a 0 if the character has not appeared, you
can see this number initialized as 32 0 bits:
00000000 00000000 00000000 00000000
*/
int checker = 0;
//loop through each String character
for (int i = 0; i < str.length(); ++i) {
/*
a through z in ASCII are charactets numbered 97 through 122, 26 characters total
with this, you get a number between 0 and 25 to represent each character index
0 for 'a' and 25 for 'z'
renamed 'val' as 'characterIndex' to be more descriptive
*/
int characterIndex = str.charAt(i) - 'a'; //char 'a' would get 0 and char 'z' would get 26
/*
created a new variable to make things clearer 'singleBitOnPosition'
It is used to calculate a number that represents the bit value of having that
character index as a 1 and the rest as a 0, this is achieved
by getting the single digit 1 and shifting it to the left as many
times as the character index requires
e.g. character 'd'
00000000 00000000 00000000 00000001
Shift 3 spaces to the left (<<) because 'd' index is number 3
1 shift: 00000000 00000000 00000000 00000010
2 shift: 00000000 00000000 00000000 00000100
3 shift: 00000000 00000000 00000000 00001000
Therefore the number representing 'd' is
00000000 00000000 00000000 00001000
*/
int singleBitOnPosition = 1 << characterIndex;
/*
This peforms an AND between the checker, which is the bit array
containing everything that has been found before and the number
representing the bit that will be turned on for this particular
character. e.g.
if we have already seen 'a', 'b' and 'd', checker will have:
checker = 00000000 00000000 00000000 00001011
And if we see 'b' again:
'b' = 00000000 00000000 00000000 00000010
it will do the following:
00000000 00000000 00000000 00001011
& (AND)
00000000 00000000 00000000 00000010
-----------------------------------
00000000 00000000 00000000 00000010
Since this number is different than '0' it means that the character
was seen before, because on that character index we already have a
1 bit value
*/
if ((checker & singleBitOnPosition) > 0) {
return false;
}
/*
Remember that
checker |= singleBitOnPosition is the same as
checker = checker | singleBitOnPosition
Sometimes it is easier to see it expanded like that.
What this achieves is that it builds the checker to have the new
value it hasnt seen, by doing an OR between checker and the value
representing this character index as a 1. e.g.
If the character is 'f' and the checker has seen 'g' and 'a', the
following will happen
'f' = 00000000 00000000 00000000 00100000
checker(seen 'a' and 'g' so far) = 00000000 00000000 00000000 01000001
00000000 00000000 00000000 00100000
| (OR)
00000000 00000000 00000000 01000001
-----------------------------------
00000000 00000000 00000000 01100001
Therefore getting a new checker as 00000000 00000000 00000000 01100001
*/
checker |= singleBitOnPosition;
}
return true;
}
我还假设您的示例来自《破解代码访谈》一书，而我的回答与此上下文有关。
为了使用这个算法来解决这个问题，我们不得不承认我们只会将字符从 a 传递到 z（小写）。
由于只有 26 个字母，并且这些字母在我们使用的编码表中进行了正确排序，这保证了我们所有的潜在差异str.charAt(i) - 'a'都将低于 32（int 变量的大小checker）。
正如 Snowbear 所解释的，我们即将将该checker变量用作位数组。让我们举个例子：
比方说
str equals "test"
检查器 == 0 (00000000000000000000000000000000)
In ASCII, val = str.charAt(i) - 'a' = 116 - 97 = 19
What about 1 << val ?
1 == 00000000000000000000000000000001
1 << 19 == 00000000000010000000000000000000
checker |= (1 << val) means checker = checker | (1 << val)
so checker = 00000000000000000000000000000000 | 00000000000010000000000000000000
checker == 524288 (00000000000010000000000000000000)
检查器 == 524288 (00000000000010000000000000000000)
val = 101 - 97 = 4
1 == 00000000000000000000000000000001
1 << 4 == 00000000000000000000000000010000
checker |= (1 << val)
so checker = 00000000000010000000000000000000 | 00000000000000000000000000010000
checker == 524304 (00000000000010000000000000010000)
依此类推..直到我们通过条件在检查器中找到特定字符的已设置位
(checker & (1 << val)) > 0
希望能帮助到你
阅读上面伊万的回答确实对我有所帮助，尽管我会有所不同。
<<in是(1 << val)移位运算符。它需要1（在二进制中表示为000000001，前面有任意数量的零 / 由内存分配）并将其向左移动val空格。由于我们只假设 az 并且每次都减去a，每个字母的值将是 0-25，这将是该字母在checker整数布尔表示中从右侧开始的索引，因为我们将按时间1向左移动checker val。
在每次检查结束时，我们都会看到|=操作员。这将合并两个二进制数，如果 a存在于该索引处的任一操作数中，则将所有0'替换为 '。在这里，这意味着无论 a存在于何处，都将被复制到中，而所有现有的 1 都将被保留。111(1 << val)1checkerchecker
正如您可能猜到的那样，1这里的 a 函数作为 true 的布尔标志。当我们检查一个字符是否已经在字符串中表示时，我们比较checker，此时它本质上是一个布尔标志（1值）数组，位于已经表示的字符的索引处，本质上是一个数组1在当前字符的索引处带有标志的布尔值。
操作员完成此&检查。与 类似，|=仅当两个操作数在该索引处都有 a 时，&运算符才会复制 a 。因此，本质上，只有已经存在的标志也会被复制。在这种情况下，这意味着只有当当前字符已经被表示时，才会在结果中的任何地方出现。如果 a出现在该操作的结果中的任何位置，则返回的布尔值的值为，并且该方法返回 false。1 1checker(1 << val)1checker & (1 << val)1> 0
我猜这就是为什么位向量也被称为位数组。因为，即使它们不是数组数据类型，也可以像使用数组一样使用它们来存储布尔标志。
上面已经提供了几个很好的答案。所以我不想重复已经说过的一切。但是确实想添加一些东西来帮助上述程序，因为我刚刚完成了同一个程序并且有几个问题但是花了一些时间之后，我对这个程序有了更清晰的了解。
首先，“检查器”用于跟踪字符串中已经遍历的字符，以查看是否有任何字符重复。
现在“checker”是一个 int 数据类型，所以它只能有 32 位或 4 个字节（取决于平台），所以这个程序只能在 32 个字符范围内的字符集上正常工作。这就是原因，该程序从每个字符中减去“a”，以使该程序仅针对小写字符运行。但是，如果您混合使用小写和大写字符，则它将不起作用。
顺便说一句，如果您不从每个字符中减去“a”（请参见下面的语句），那么该程序将仅适用于带有大写字符的字符串或仅带有小写字符的字符串。所以上述程序的范围也从小写字符增加到大写字符，但它们不能混合在一起。
int val = str.charAt(i) - 'a';
但是我想使用按位运算编写一个通用程序，它应该适用于任何 ASCII 字符，而不必担心大写、小写、数字或任何特殊字符。为了做到这一点，我们的“检查器”应该足够大以存储 256 个字符（ASCII 字符集大小）。但是 Java 中的 int 不能工作，因为它只能存储 32 位。因此，在下面的程序中，我使用了 JDK 中可用的 BitSet 类，它可以在实例化 BitSet 对象时传递任何用户定义的大小。
这是一个与上面使用位运算符编写的程序执行相同操作的程序，但该程序适用于具有 ASCII 字符集中任何字符的字符串。
public static boolean isUniqueStringUsingBitVectorClass(String s) {
final int ASCII_CHARACTER_SET_SIZE = 256;
final BitSet tracker = new BitSet(ASCII_CHARACTER_SET_SIZE);
// if more than 256 ASCII characters then there can't be unique characters
if(s.length() > 256) {
return false;
}
//this will be used to keep the location of each character in String
final BitSet charBitLocation = new BitSet(ASCII_CHARACTER_SET_SIZE);
for(int i = 0; i < s.length(); i++) {
int charVal = s.charAt(i);
charBitLocation.set(charVal); //set the char location in BitSet
//check if tracker has already bit set with the bit present in charBitLocation
if(tracker.intersects(charBitLocation)) {
return false;
}
//set the tracker with new bit from charBitLocation
tracker.or(charBitLocation);
charBitLocation.clear(); //clear charBitLocation to store bit for character in the next iteration of the loop
}
return true;
}
简单说明（下面有JS代码）
32-bitDEC64对于 JS。0th，如果我们a在字符串中找到，我们设置索引， 1stfor b& 依此类推。操作总结：
checkerindexInt-32-Arraysif操作的输出是1output == 1
checker变量在两个数组中都设置了特定的索引位output == 0
checkerindex1checker假设：
a97下面给出的是JavaScript源代码。
function checkIfUniqueChars (str) {
var checker = 0; // 32 or 64 bit integer variable
for (var i = 0; i< str.length; i++) {
var index = str[i].charCodeAt(0) - 96;
var bitRepresentationOfIndex = 1 << index;
if ( (checker & bitRepresentationOfIndex) > 1) {
console.log(str, false);
return false;
} else {
checker = (checker | bitRepresentationOfIndex);
}
}
console.log(str, true);
return true;
}
checkIfUniqueChars("abcdefghi"); // true
checkIfUniqueChars("aabcdefghi"); // false
checkIfUniqueChars("abbcdefghi"); // false
checkIfUniqueChars("abcdefghii"); // false
checkIfUniqueChars("abcdefghii"); // false
请注意，在 JS 中，尽管整数是 64 位的，但按位运算总是在 32 位上完成。
示例：
如果字符串是aathen：
// checker is intialized to 32-bit-Int(0)
// therefore, checker is
checker= 00000000000000000000000000000000
我 = 0
str[0] is 'a'
str[i].charCodeAt(0) - 96 = 1
checker 'AND' 32-bit-Int(1) = 00000000000000000000000000000000
Boolean(0) == false
// So, we go for the '`OR`' operation.
checker = checker OR 32-bit-Int(1)
checker = 00000000000000000000000000000001
我 = 1
str[1] is 'a'
str[i].charCodeAt(0) - 96 = 1
checker= 00000000000000000000000000000001
a = 00000000000000000000000000000001
checker 'AND' 32-bit-Int(1) = 00000000000000000000000000000001
Boolean(1) == true
// We've our duplicate now
让我们逐行分解代码。
整数检查器 = 0; 我们正在启动一个检查器，它将帮助我们找到重复值。
int val = str.charAt(i) - 'a'; 我们在字符串的第 i 个位置获取字符的 ASCII 值，然后用 'a' 的 ASCII 值减去它。由于假设字符串仅为低位字符，因此字符数限制为 26。因此，'val' 的值将始终 >= 0。
if ((checker & (1 << val)) > 0) 返回 false；
检查器 |= (1 << val);
现在这是棘手的部分。让我们考虑一个带有字符串“abcda”的示例。理想情况下，这应该返回 false。
对于循环迭代 1：
检查器：000000000000000000000000000000000
值：97-97 = 0
1 << 0: 00000000000000000000000000000001
检查器 & (1 << val): 00000000000000000000000000000000 不是 > 0
因此检查器：000000000000000000000000000000001
对于循环迭代 2：
检查器：00000000000000000000000000000001
值：98-97 = 1
1 << 1: 00000000000000000000000000000010
检查器 & (1 << val): 00000000000000000000000000000000 不是 > 0
因此检查器：00000000000000000000000000000011
对于循环迭代 3：
检查器：00000000000000000000000000000011
值：99-97 = 2
1 << 2: 00000000000000000000000000000100
检查器 & (1 << val): 00000000000000000000000000000000 不是 > 0
因此检查器：00000000000000000000000000000111
对于循环迭代 4：
检查器：00000000000000000000000000000111
值：100-97 = 3
1 << 3: 00000000000000000000000000001000
检查器 & (1 << val): 00000000000000000000000000000000 不是 > 0
因此检查器：00000000000000000000000000001111
对于循环迭代 5：
检查器：000000000000000000000000000001111
值：97-97 = 0
1 << 0: 00000000000000000000000000000001
检查器 & (1 << val): 00000000000000000000000000000001 > 0
因此返回false。
public static void main (String[] args)
{
//In order to understand this algorithm, it is necessary to understand the following:
//int checker = 0;
//Here we are using the primitive int almost like an array of size 32 where the only values can be 1 or 0
//Since in Java, we have 4 bytes per int, 8 bits per byte, we have a total of 4x8=32 bits to work with
//int val = str.charAt(i) - 'a';
//In order to understand what is going on here, we must realize that all characters have a numeric value
for (int i = 0; i < 256; i++)
{
char val = (char)i;
System.out.print(val);
}
//The output is something like:
// !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~ ¡¢£¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæçèéêëìíîïðñòóôõö÷øùúûüýþÿ
//There seems to be ~15 leading spaces that do not copy paste well, so I had to use real spaces instead
//To only print the characters from 'a' on forward:
System.out.println();
System.out.println();
for (int i=0; i < 256; i++)
{
char val = (char)i;
//char val2 = val + 'a'; //incompatible types. required: char found: int
int val2 = val + 'a'; //shift to the 'a', we must use an int here otherwise the compiler will complain
char val3 = (char)val2; //convert back to char. there should be a more elegant way of doing this.
System.out.print(val3);
}
//Notice how the following does not work:
System.out.println();
System.out.println();
for (int i=0; i < 256; i++)
{
char val = (char)i;
int val2 = val - 'a';
char val3 = (char)val2;
System.out.print(val3);
}
//I'm not sure why this spills out into 2 lines:
//EDIT I cant seem to copy this into stackoverflow!
System.out.println();
System.out.println();
//So back to our original algorithm:
//int val = str.charAt(i) - 'a';
//We convert the i'th character of the String to a character, and shift it to the right, since adding shifts to the right and subtracting shifts to the left it seems
//if ((checker & (1 << val)) > 0) return false;
//This line is quite a mouthful, lets break it down:
System.out.println(0<<0);
//00000000000000000000000000000000
System.out.println(0<<1);
//00000000000000000000000000000000
System.out.println(0<<2);
//00000000000000000000000000000000
System.out.println(0<<3);
//00000000000000000000000000000000
System.out.println(1<<0);
//00000000000000000000000000000001
System.out.println(1<<1);
//00000000000000000000000000000010 == 2
System.out.println(1<<2);
//00000000000000000000000000000100 == 4
System.out.println(1<<3);
//00000000000000000000000000001000 == 8
System.out.println(2<<0);
//00000000000000000000000000000010 == 2
System.out.println(2<<1);
//00000000000000000000000000000100 == 4
System.out.println(2<<2);
// == 8
System.out.println(2<<3);
// == 16
System.out.println("3<<0 == "+(3<<0));
// != 4 why 3???
System.out.println(3<<1);
//00000000000000000000000000000011 == 3
//shift left by 1
//00000000000000000000000000000110 == 6
System.out.println(3<<2);
//00000000000000000000000000000011 == 3
//shift left by 2
//00000000000000000000000000001100 == 12
System.out.println(3<<3);
// 24
//It seems that the - 'a' is not necessary
//Back to if ((checker & (1 << val)) > 0) return false;
//(1 << val means we simply shift 1 by the numeric representation of the current character
//the bitwise & works as such:
System.out.println();
System.out.println();
System.out.println(0&0); //0
System.out.println(0&1); //0
System.out.println(0&2); //0
System.out.println();
System.out.println();
System.out.println(1&0); //0
System.out.println(1&1); //1
System.out.println(1&2); //0
System.out.println(1&3); //1
System.out.println();
System.out.println();
System.out.println(2&0); //0
System.out.println(2&1); //0 0010 & 0001 == 0000 = 0
System.out.println(2&2); //2 0010 & 0010 == 2
System.out.println(2&3); //2 0010 & 0011 = 0010 == 2
System.out.println();
System.out.println();
System.out.println(3&0); //0 0011 & 0000 == 0
System.out.println(3&1); //1 0011 & 0001 == 0001 == 1
System.out.println(3&2); //2 0011 & 0010 == 0010 == 2, 0&1 = 0 1&1 = 1
System.out.println(3&3); //3 why?? 3 == 0011 & 0011 == 3???
System.out.println(9&11); // should be... 1001 & 1011 == 1001 == 8+1 == 9?? yay!
//so when we do (1 << val), we take 0001 and shift it by say, 97 for 'a', since any 'a' is also 97
//why is it that the result of bitwise & is > 0 means its a dupe?
//lets see..
//0011 & 0011 is 0011 means its a dupe
//0000 & 0011 is 0000 means no dupe
//0010 & 0001 is 0011 means its no dupe
//hmm
//only when it is all 0000 means its no dupe
//so moving on:
//checker |= (1 << val)
//the |= needs exploring:
int x = 0;
int y = 1;
int z = 2;
int a = 3;
int b = 4;
System.out.println("x|=1 "+(x|=1)); //1
System.out.println(x|=1); //1
System.out.println(x|=1); //1
System.out.println(x|=1); //1
System.out.println(x|=1); //1
System.out.println(y|=1); // 0001 |= 0001 == ?? 1????
System.out.println(y|=2); // ??? == 3 why??? 0001 |= 0010 == 3... hmm
System.out.println(y); //should be 3??
System.out.println(y|=1); //already 3 so... 0011 |= 0001... maybe 0011 again? 3?
System.out.println(y|=2); //0011 |= 0010..... hmm maybe.. 0011??? still 3? yup!
System.out.println(y|=3); //0011 |= 0011, still 3
System.out.println(y|=4); //0011 |= 0100.. should be... 0111? so... 11? no its 7
System.out.println(y|=5); //so we're at 7 which is 0111, 0111 |= 0101 means 0111 still 7
System.out.println(b|=9); //so 0100 |= 1001 is... seems like xor?? or just or i think, just or... so its 1101 so its 13? YAY!
//so the |= is just a bitwise OR!
}
public static boolean isUniqueChars(String str) {
int checker = 0;
for (int i = 0; i < str.length(); ++i) {
int val = str.charAt(i) - 'a'; //the - 'a' is just smoke and mirrors! not necessary!
if ((checker & (1 << val)) > 0) return false;
checker |= (1 << val);
}
return true;
}
public static boolean is_unique(String input)
{
int using_int_as_32_flags = 0;
for (int i=0; i < input.length(); i++)
{
int numeric_representation_of_char_at_i = input.charAt(i);
int using_0001_and_shifting_it_by_the_numeric_representation = 1 << numeric_representation_of_char_at_i; //here we shift the bitwise representation of 1 by the numeric val of the character
int result_of_bitwise_and = using_int_as_32_flags & using_0001_and_shifting_it_by_the_numeric_representation;
boolean already_bit_flagged = result_of_bitwise_and > 0; //needs clarification why is it that the result of bitwise & is > 0 means its a dupe?
if (already_bit_flagged)
return false;
using_int_as_32_flags |= using_0001_and_shifting_it_by_the_numeric_representation;
}
return true;
}
以前的帖子很好地解释了代码块的作用，我想使用 BitSet java 数据结构添加我的简单解决方案：
private static String isUniqueCharsUsingBitSet(String string) {
BitSet bitSet =new BitSet();
for (int i = 0; i < string.length(); ++i) {
int val = string.charAt(i);
if(bitSet.get(val)) return "NO";
bitSet.set(val);
}
return "YES";
}
Line 1: public static boolean isUniqueChars(String str) {
Line 2: int checker = 0;
Line 3: for (int i = 0; i < str.length(); ++i) {
Line 4: int val = str.charAt(i) - 'a';
Line 5: if ((checker & (1 << val)) > 0) return false;
Line 6: checker |= (1 << val);
Line 7: }
Line 8: return true;
Line 9: }
我理解使用 Javascript 的方式。假设输入var inputChar = "abca"; //find if inputChar has all unique characters
Line 4: int val = str.charAt(i) - 'a';
上面的行查找 inputChar 中第一个字符的二进制值，即a，a = 97 in ascii，然后将 97 转换为二进制变为 1100001。
在 Javascript 例如："a".charCodeAt().toString(2) 返回 1100001
checker = 0// 二进制 32 位表示 = 0000000000000000000000000
checker = 1100001 | checker;//checker变成1100001（在32位表示中它变成000000000.....00001100001）
但我希望我的位掩码（int checker）只设置一位，但检查器是 1100001
Line 4: int val = str.charAt(i) - 'a';
现在上面的代码就派上用场了。我总是减去 97（a 的 ASCII 值）
val = 0; // 97 - 97 Which is a - a
val = 1; // 98 - 97 Which is b - a
val = 1; // 99 - 97 Which is c - a
让我们使用val哪个被重置
第 5 行和第 6 行很好解释 @Ivan 答案
以防万一有人正在使用位向量在字符串中寻找与唯一字符等效的 kotlin
fun isUnique(str: String): Boolean {
var checker = 0
for (i in str.indices) {
val bit = str.get(i) - 'a'
if (checker.and(1 shl bit) > 0) return false
checker = checker.or(1 shl bit)
}
return true
}