方法应输入一个单词,然后以递归方式遍历字符串,然后查找与字母表两端距离相同的字母并将其删除。如果删除匹配项,则这些字母不能再次使用。如果每个字母都被删除,则表示匹配。

for (int i = 1; i < word.length()-1; i++) { if (word.charAt(0) + word.charAt(i) == 155) { StringBuilder sb = new StringBuilder(word); sb.deleteCharAt(0); sb.deleteCharAt(i); String strNew = sb.toString(); System.out.println(strNew); return isAlphaOpp(strNew); } } return false; }

最佳答案

我对您的方法做了一些修改,请看一下。如果您的字符串全部为大写字母,则需要与155进行比较;如果所有小写字母都需要与219相比较。如@Raghu所建议的,这不需要递归(这会使事情变得复杂),但是我假设您想尝试使用递归。

public static boolean isAlphaOpp (String word)
    {
        //if word has odd number of characters, it cannot be an alpha opp
        if (word.length() % 2 != 0)
        {
            return false;
        }
        //if string makes it to 0, then word must be an alpha opp
        if (word.length() == 0)
        {
            return true;
        }

        /*if (word.charAt(0) + word.charAt(word.length()-1) == 155)
            {
                System.out.println(word.substring(1, word.length()-1));
                return isAlphaOpp(word.substring(1, word.length()-1));
            }
        */
        //Should go thru each letter and compare the values with char(0). If char(0) +     //char(i) == 155 (a match) then it should remove them and call the method again.
        int length = word.length()-1;
        int start = 0;
        String newStr = null;
        while(start < length) {

            if(word.charAt(start) + word.charAt(length) == 219) {
                StringBuilder sb = new StringBuilder(word);
                sb.deleteCharAt(length);
                sb.deleteCharAt(start);
                newStr = sb.toString();
                System.out.println(newStr);
                start++;
                length--;
                break;
            } else {
                start++;
            }
        }
        if(newStr != null) {
            return isAlphaOpp(newStr);
        }
        return false;
    }

08-27 07:29
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