您可以为此使用按位运算:

You can use bitwise operations for that:

private static boolean isPangrams(String ip) {
    int flags = 0;
    for(char current : ip.toLowerCase().toCharArray()) {
        if (current >= 'a' && current <= 'z') {
            flags |= 0x01<<(current-'a');
        }
    }
    return flags == 0x3ffffff;
}

jDoodle

代码的工作方式如下:我们认为一个int是32位数字.直到26位的每个位都是一个标志(可以说是boolean).最初，所有标志都是false，因为我们用0初始化了flags.

The code works as follows: we consider an int which is a 32-bit number. Each bit up to 26 is a flag (a boolean so to speak). Initially all flags are false because we initialize flags with 0.

现在，我们遍历字符串的字符.如果字符是小写字母，则将相应标志的标志设置为true(无论之前是否已将其设置为true).

Now we iterate over the characters of the string. In case the character is a lowercase letter, we set the flag of the corresponding flag to true (regardless whether it has been set to true before).

最后，我们检查最低的26位是否都设置为true.如果是，则flags等于0x3ffffff(这是一个等于1111111111111111111111二进制的十六进制数.如果是，则返回true.否则返回false.

Finally we inspect whether the lowest 26 bits are all set to true. If so, flags is equal to 0x3ffffff (which is a hexadecimal number equal to 1111111111111111111111 binary. If so we return true. Otherwise we return false.

通常，按位运算比if语句和布尔值更快，因此我希望该程序要快得多.

Usually bitwise operations are faster than if statements and booleans so I expect this program to be a significant bit faster.