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;
}