java - 字符串和 2 个字母

Question

您好，我是一名学生，我的问题更详细是这样的。

Given a string and two letters (c1 and c2), return a count of the number of times "axb" occurs in the string, where x is any character. For example, given the string "antiaircraft" and the letters 'a' and 't', your method should return 2. The three-letter patterns may overlap. For example, "aaaa" has two occurrences of "axa".

到目前为止我所写的是

public int countAxA(String str, char c1, char c2) {

 int count = 0;
    for (int i=0; i < str.length(); i++)
    {
        if (str.charAt(i) == c2)
        {
             count++;
        }
    }
    return count;
}

因此，根据我的作业，某些输入有效，但其他输入无效。我有什么遗漏的吗？

Answer 1

只需迭代一次并跟踪，如果您碰巧找到第一个字符，则进行前瞻:

public int countAxA(String s, char one, char two) {
    char[] cs = s.toCharArray();
    int count = 0;
    for (int i = 0; i < cs.length - 2; i++) { //don't need to go beyond 3rd last char
        if (cs[i] == one && cs[i + 2] == two) {
            count++;
        }
    }
    return count;
}