java - 如何在没有任何内置函数的情况下从java中的两个字符串中删除重复的字母表？

Question

这里我有两个字符串。由此，如何删除常见的字母/字符并存储在结果(第三个)字符串中？

<%
    String[] firstString = {"google out"};
    String[] secondString = {"stack overflow"};
    String[] result={""};
        for (int i = 0,k=0; i < firstString.length; i++,k++) {
            for (int j = 0; j < secondString.length; j++) {
                if (firstString[i].equalsIgnoreCase(secondString[j])) {
            } else {
                result[k]=result[j]+firstString[i];
                out.println(result[k]);
            }
        }
    }
%>

预期结果是:
g l e o u t s c k v f w

Answer 1

这是一种方法，

// Write a static helper method.
public static boolean contains(char[] in, int index, char t) {
    for (int i = 0; i < index; i++) {
        if (in[i] == t) return true;
    }
    return false;
}

public static void main(String[] args) {
    String firstString = "google out"; // String(s) not String arrays
    String secondString = "stack overflow";
    // output cannot be larger then the sum of the inputs.
    char[] out = new char[firstString.length() + secondString.length()];
    int index = 0;
    // Add all unique chars from firstString
    for (char c : firstString.toCharArray()) {
        if (! contains(out, index, c)) {
            out[index++] = c;
        }
    }
    // Add all unique chars from secondString
    for (char c : secondString.toCharArray()) {
        if (! contains(out, index, c)) {
            out[index++] = c;
        }
    }
    // Create a correctly sized output array.
    char[] s = new char[index];
    for (int i = 0; i < index; i++) {
        s[i] = out[i];
    }
    // Just print it.
    System.out.println(Arrays.toString(s));
}

输出是

[g, o, l, e,  , u, t, s, a, c, k, v, r, f, w]

编辑

如果您的预期输出不正确，而您实际上想要出现在两个字符串中的字符。

public static void main(String[] args) {
    String firstString = "google out";
    String secondString = "stack overflow";
    char[] out = new char[firstString.length() + secondString.length()];
    int index = 0;
    for (char c : firstString.toCharArray()) {
        if (contains(secondString.toCharArray(), secondString.length(), c)) {
            out[index++] = c;
        }
    }
    char[] s = new char[index];
    for (int i = 0; i < index; i++) {
        s[i] = out[i];
    }
    System.out.println(Arrays.toString(s));
}

哪些输出

[o, o, l, e,  , o, t]

编辑2

如果您真的想要与之相反，请将调用更改为 contains 并添加第二个循环(用于反向关系)-

    for (char c : firstString.toCharArray()) {
        if (! contains(secondString.toCharArray(), secondString.length(), c)) {
            out[index++] = c;
        }
    }
    for (char c : secondString.toCharArray()) {
        if (! contains(firstString.toCharArray(), firstString.length(), c)) {
            out[index++] = c;
        }
    }

然后会输出

[g, g, u, s, a, c, k, v, r, f, w]