java.lang.StringIndexOutOfBounds improvement

Question

public class DuplicateLetters {

// CW2.2 Lab-Group-06 Question 5
// You are given two non-empty strings source and target, all lowercase letters, from a user.
// The user intends to type the string given in target.
// However, due to using an old keyboard,
// when the user was typing, a number characters may be duplicated, and typed one or more times.
// The source is what the user actually typed, which may or may not be the intended target.
// Return true if and only if it is possible that some characters of target have been duplicated in source,
// while the user intended to type the target string.
// You must use String methods in lecture notes.
// You must NOT use StringBuilder or Regular Expression methods.

public static boolean duplicateLetters(String source, String target) {
boolean test = false;

for(int i=0; i<source.length(); i++) {
for(int j=i; j<target.length(); j++) {
if(Character.toString(source.charAt(i)).equals(Character.toString(target.charAt(j)))) {
test = true;
}
else {
test = false;
j--;    
}
i++;
}
}
return test;
}

public static void main(String[] args) {

System.out.println(duplicateLetters("andddrew", "andrew"));
// true, there are duplicated letters typed

System.out.println(duplicateLetters("lllejiiee", "leejie"));
// false, note the initial letter e
// there is only one initial e in the source, whereas there is a double initial ee in the target

System.out.println(duplicateLetters("cherry", "cherry"));
// true, no duplicated letters typed this time
}

这是我针对这个问题的代码，但它一直得到java.lang.StringIndexOutOfBounds错误。所以，我想知道我的代码有什么问题，以及如何改进。

Answer 1

这是一个正确的运行版本。请注意，您还需要检查不匹配的新字符是否与目标的最后一个字符匹配。否则CCD_ 1将失败。(这仍然会发生在鲁胡尔已经提出的指数修复中(。

public static boolean duplicateLetters(String source, String target) {
int i = 0;
char lastTargetChar = target.charAt(0);
for(int j=0; j<target.length(); j++) {
while ( source.charAt(i) != target.charAt(j)) {
// we only go in here if the char in source is wrong, thus assume a dupe, but check if it's a dupe!
if (lastTargetChar != source.charAt(i)) {
return false;
}
i++;
if (i >= source.length()) {
return false;
}
}
// if we end here, source char was found ok, so remember last char and go to next char in source
lastTargetChar = target.charAt(j);
i++;
}
return true;
}

优化：

1. 在目标上循环，以便更容易地检查目标中的双字母(例如"letter"one_answers"lettttter"(
2. 出现问题时立即退出循环。你不需要继续，因为结果在任何情况下都是正确的

Answer 2

当i的值超过发生的source.length异常时，您正在内部循环中递增值i。

for(int j=i; j<target.length(); j++) {
if(Character.toString(source.charAt(i)).equals(Character.toString(target.charAt(j)))) {
test = true;
}
else {
test = false;
j--;    // you don't need to decrement the value of j.
}
i++; // problematic code.
}

要解决此问题，请在比较或中断循环之前添加一个检查。例如：

...
i++; // break the loop so that i remains consistent.
if(i>=source.length()) break;

此外，您的逻辑过于复杂和错误。由于修复这些错误超出了范围，所以你应该自己找(作业(：P