你好,我不能做这个工作,如果这个词包含在主词中,我会得到一个主词,然后是另一个子词,应该删除该部分。
//Example
//fmrog (in.nextLine)(main word)
//4 (in.nextInt)(the amount of sub words)
//roc(in.nextLine)(not contained)
//gor(in.nextLine)(not contained)
//rog(in.nextLine)(contained)
//ogr(in.nextLine)(not contained)
//result:fm
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
StringBuilder mainWord = new StringBuilder(in.nextLine);
int n = in.nextInt();
StringBuilder MainWord2 = new StringBuilder(mainWord);
in.nextLine();
for (int i = 0; i < n; i++) {
String subWord = in.nextLine();
int chars = subWord.length();
if (chars> mainWord.length()){
continue;
}
for (int j = 0; j < subWord.length(); j++) {
int r = 0;
for (int k = 0; k < mainWord.length(); k++) {
r++;
if (k > MainWord2.length() - 1) {
break;
}
if (MainWord2.charAt(k) == subWord.charAt(j)) {
break;
}
}
if (r <= MainWord2.length() && MainWord2.charAt(r-1) == subWord.charAt(j)) {
MainWord2.deleteCharAt(r - 1);
if (j >= subWord.length() -1 ) {
mainWord = MainWord2;
break;
}
}
if (r > MainWord2.length()) {
MainWord2 = mainWord;
break;
}
}
}
System.out.println(mainWord);
}
}
老实说,我很困惑,也许有一种更简单的方法可以解决这个问题。最主要的是,当我写一个案例时:"超级 2 pe surr"在"surr"结束时,当我删除聊天时,两个 StringBuilder 开始充当一个,其中一个也发生了变化
没必要让它变得这么复杂。
String input = // complete user input
String[] words = String.split(input);
String mainWord = words[0];
int numWords = Integer.parseInt(words[1]); // this variable isn't needed
for(int i = 2; i < words.length; i++) {
if (mainWord contains words[i]) {
mainWord = mainWord.replace(words[i], ""); // remove subword from mainword
}
}
最后,mainWord 将是原始的主词,没有任何后来输入的子词。
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
in.nextLine();
String[] words = new String[n];
for (int i = 0; i <n ; i++) {
words[i] = in.nextLine();
}
String mainWord = words[0];
for (int i = 1; i <words.length ; i++) {
if (mainWord.contains(words[i])){
mainWord = mainWord.replace(words[i], "");
}
}
System.out.println(mainWord);
}
}
在这里,但问题是,如果字母不是一个相邻的字母,它不会删除子词。