Java String indexOf 忽略重复字符

Question

我需要 java.lang.String.indexOf(String str, int fromIndex) 的一个版本，它可以忽略重复字符，因为它返回字母第一次出现的索引。

这与 String 类的 indexOf 方法不同，因为它要求字符串只是字符串的单个实例，因此如果字符串连续出现两次或多次，它将不会返回任何这些值。如果没有单个实例，该方法应该像普通的indexOf一样返回-1。

例如，我们将调用方法indexOfNoDup:

static String string = "aabaccbdd";

public static void main(String[] args)
{
    indexOfNoDup("a", 0);
    //Output: 3
    //The first single occurrence of "a" is in position 3. The "a"s in position 0 and 1 are not returned because they are not single instances.

    indexOfNoDup("b", 4);
    //Output: 6
    //The first single occurrence of "b" at or after position 4 is in position 6.

    indexOfNoDup("c", 0);
    //Output: -1
    //There is no single instance of "c" in the line.

    indexOfNoDup("a", 1);
    //Output: 1
    //The first single occurrence of "a" at or after position 1 is in position 1.
}

Answer 1

有时，首先对您的问题进行建模会有所帮助。我们可以使用有限状态机对此进行建模。

^{使用 draw.io 创建的图像}

所以我们想要做的是实现该状态机。有几种方法可以做到这一点。这是一种这样的方法(假设我正确实现了它，但这应该不会太难调试):

int index = -1;
int state = 0;
while (index < string.length()) {
    int new_index = string.indexOf(value, index + 1);
    if (new_index == -1) {
        switch (state) {
            case 0: return -1;
            case 1: return index;
            case 2: return -1;
        }
    }
    if (new_index == index + value.length()) {
        if (state < 2) state++;
    } else {
        state = 0;
    }
    index = new_index;
}

对于单个字符的具体情况，我们可以这样做:

int index = -1;
int found_index = -1;
int state = 0;
while (index < string.length()) {
    if (string.charAt(index + 1) == value) { //assuming value is type char, else do value.charAt(0)
        if (state < 2) state++;
        if (state == 1) found_index = index;
    } else {
        state = 0;
    }
    index++;
}
switch (state) {
    case 0: return -1;
    case 1: return found_index;
    case 2: return -1;
}