如果我有2个句子:
String code1 = "XXXji X9 XX22 Xs dXXXd i l9XX bXX45.";
String code2 = "hello my name is david i like bread.";
基本上,一个加扰而一个不加(每个单词具有相同数量的字符)
我将如何得出这样的结论:
String answer = "XXXlo Xy XXme Xs dXXXd i liXX bXXad.";
我会有code1
和code2
方便,我不是要解密任何东西,只是整合字符串。
您可以尝试使用StringBuilder,如下所示:
public class StringConverter {
public static void main(String[] args) {
String code1 = "XXXji X9 XX22 Xs dXXXd i l9XX bXX45.";
String code2 = "hello my name is david i like bread.";
StringBuilder result = new StringBuilder(code2);
for(int i = 0; i < code1.length(); i++) {
char c1 = code1.charAt(i);
if (c1 == 'X') {
result.replace(i, i+1, "X");
}
}
System.out.println("String answer is");
System.out.println(result);
}
}
产量
String answer is
XXXlo Xy XXme Xs dXXXd i liXX bXXad.
希望这个解决方案适合您。
此代码应返回您的要求:
char[] template = code1.toCharArray();
char[] content = code1.toCharArray();
if(template.length == content.length){
char[] newArray = new char[template.length];
for(int i = 0; i < template.length; i++){
if(template[i] != 'X'){
newArray[i] = content[i];
} else{
newArray[i] = 'X';
}
}
return newArray;
任何其他实用程序只会使用类似于此算法的东西