带移位和运算符的独特字符：不明白这段代码 [英] unique chars with shift and operators : don't understand this code

问题描述

我不理解循环中的行：我们取字符并减去 a ，所以10？ （为什么？）

然后 1<< val ：我们将val换成1？ （为什么？）

和检查器是0，那么我们如何达到> 0 条件？

i don't understand the lines in the loop : we take the character and subtract a, so "10" ? (why?)
then 1 << val : we shift 1 by val ? (why?)
and checker is 0, so how do we reach > 0 in the condition?

    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;
}

谢谢

推荐答案

该代码假定 str 由小写字母组成，如果没有重复字母则返回true。它的工作方式如下：

The code assumes that str is made of lower case letters and returns true if there are no repeating letters. It works like this:

checker 用作位图，即使用此变量中的每个位跟踪一个字母。从字符中减去'a'为'a'提供0，为'c'提供1，为'c'提供2等等。将此数字向左移1表示'a'为1，'b'为2，'c为4 '等。

checker is used as a bitmap, that is, each bit in this variable is used to keep track of one letter. Subtracting 'a' from the character gives 0 for 'a', 1 for 'b', 2 for 'c' etc. Shifting 1 left by this number gives 1 for 'a', 2 for 'b', 4 for 'c' etc.

通过oring（ | ）这个值与 checker 代码跟踪以前遇到的字符。所以，如果我们遇到第二个'a'，例如， checker 它的第一个位设置（由& 在if语句中），所以我们知道 str 有重复。

By oring (|) this value with checker the code keeps track of previously encountered characters. So, if we encounter a second 'a' for example, checker has it's first bit set (which is tested by & in the if statement), so we know that str has duplicates.

简而言之， checker 用作更快更紧凑的bool数组。这种和类似的技术称为位操作。

In short, checker is used as a faster and more compact array of bool. This and similar techniques are called bit manipulation.

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