如何检查一个字符串可以使用另一个字符串中的字符拼写？

Question

Example: String a = "ACAHBBA" and String b = "ABAB" should return true, since both strings can spell ABAB.

I have tried with contains(), but that only works for equal sequences.

// The code should look like this. 
public class task10 {
    public static boolean contains(String a, String b) {
        // check if b can be spelled using characters from a. 
        // if it can. return true. 
        // else
        return false;
    }
}

Posible solution?

public static boolean contains(String a, String b) {

    for (int i = 0; i < b.length(); i++) {
        if (a.indexOf(b.charAt(i)) == -1) {
            return false;
        }
    }
    return true;
}

Answer 1

Simply iterate thru one string and get the index of the character. If >= 0, replace character with non-alphabetic character and repeat. This algorithm presumes the need to match the correct number of characters. For example, hello would return false if the character set was helo .

    public static boolean spelledFrom(String word, String chars) {
      StringBuilder sb = new StringBuilder(chars);
      for (String c : word.split("")) {
         int i;
         if ((i = sb.indexOf(c)) < 0) {
            return false;
         }
         sb.setCharAt(i, '#');
      }
      return true;
   }

Answer 2

You can try this:

public static boolean canSpell(String a, String b)
{
    String shorter = (a.length() <= b.length()) ? a : b;
    String longer = (shorter.equals(a)) ? b : a;

    for(int i = 0; i < shorter.length(); i++)
    {
        if(!longer.contains("" + shorter.charAt(i)))
            return false;
    }

    return true;
}

Once you've identified the shorter string, you just need to verify that each of its chars are contained in the longer string. This solution doesn't verify if a char "has already been used", which means inserting "AB" and "ABBA" will return true. If you need to do this, you just need to delete the verified char from the longer string in every loop.