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？

这里我有两个字符串。 从那，我怎样才能删除常见的字母/字符并存储在result（third）String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  

预期结果是：
gleoutsckvfw

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，并且您实际上想要两个字符串中出现的字符。

 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)); } 

哪个输出

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

编辑2

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

  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; } } 

然后输出

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

Answer 2

获得独特字母的最快方法可以是：

 String firstString = "google out"; String secondString = "stack overflow"; Set output = new HashSet(); Collections.addAll(output, (firstString + secondString).split("")); 

输出（包含在开头有一个空格的排序字母，因为它也是唯一字符）：

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

Answer 3

这是另一种方法：

 private String getUnicChars(String first, String second) { String result = ""; for (int i = 0; i < first.length(); i++) { if (!second.contains(String.valueOf(first.charAt(i)))) { result = result + first.charAt(i); } } return result; } public String yourMethod(String firstString, String secondString) { return getUnicChars(firstString, secondString) + getUnicChars(secondString, firstString); } 

然后你调用yourMethod("google out", "stack overflow"); 。

[Actualization]我假设您只是要求检查两个字符串之间的redondant char，而不是在一个String中。 在我的解决方案中，'g'字符将被给出两次。[/ actualization]

Answer 4

 public class RoughWork2 { // function to remove duplicates first public static String removeDuplicate(String val) { char[] chars = val.toCharArray(); Set charSet = new LinkedHashSet(); for (char c : chars) { charSet.add(c); } StringBuilder sb = new StringBuilder(); for (Character character : charSet) { sb.append(character); } return sb.toString(); } public boolean sort(String a, String b) { String arg1 = removeDuplicate(a); String arg2 = removeDuplicate(b); if (arg1.length() != arg2.length()) { System.out.print("two strings count are not equal"); return false; } else { for (int x = 0; x < arg1.length(); x++) { int c = (int) arg1.charAt(x); for (int y = 0; y < arg2.length(); y++) { int d = (int) arg2.charAt(y); if (c == d) { System.out.print(Character.toChars(c)); } else { System.out.print("*"); } } System.out.println(); } return true; } } public static void main(String[] args) { RoughWork2 rw = new RoughWork2(); rw.sort("hello", "hallo"); } }