确定字符串是否包含所有唯一字符的Java代码如何工作? [英] How does this Java code which determines whether a String contains all unique characters work?

问题描述

我有一个在网上找到的程序，该程序基本上可以告诉String是否包含所有唯一字符，下面是代码

I have a program which I found online which basically tells whether the String contains all unique characters, below is the code

private static boolean areCharsUnique(String str) {
        if (str.length() > 256)
            return false;
        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;
    }

我对这行代码if ((checker & (1 << val)) > 0)和

checker |= (1 << val);

我知道<<是左移运算符，但是在上述情况下左移到底有多大帮助呢?

I know that << is a left shift operator, but how exactly left shifting helps in the above situation?

简而言之，以上程序如何工作?

推荐答案

此代码在ASCII字符集在其映射中具有连续字符的前提下起作用，因此a == 97, b == 98，依此类推.

This code works under the assumption that ASCII character set has consecutive character in its mapping so that a == 97, b == 98, and so on.

从此开始，您可以计算与a字符(例如'e' - 'a' = 5)的距离差.此距离用于设置(通过checker |= (1 << val))整数(具有32位)中的一位.

Starting from this you can calculate a delta distance from the a character, eg 'e' - 'a' = 5. This distance is used to set (through checker |= (1 << val)) a bit in a integer number (which has 32 bits).

因此，如果找到一个'e'字符，那么checker的索引5的位将设置为1.

So if a 'e' character is found then bit at index 5 is set to 1 for checker.

通过确保您永远不会(通过if (checker & (1 << val)) > 0))找到以前已经设置的位，对每个字符进行此操作.

This is done for every char by ensuring that you never find a bit that has been already set previously (through if (checker & (1 << val)) > 0)).

这仅适用于小写字符(即使int具有32位). HashSet<Character>肯定会更好.

This works only for lowercase characters (even because int has 32 bits). An HashSet<Character> would be surely better.

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