您好,我无法完成这项工作,如果在主词中包含该词,则给我一个主词,然后再提供另一个子词,应删除该部分。
//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,我可能会遇到一种更简单的方法来解决此问题。最主要的是,当我编写类似“ Super 2 pe surr”的案例时,在“ surr”末尾,当我删除chatAt时,两个StringBuilder开始充当一个。
最佳答案
无需使其变得如此复杂。
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mainWord,不包含以后输入的任何子词。
关于java - Java从单词中删除部分,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/49702701/