我对此程序有疑问。 使用方案中，用户应输入一个数字（要输入多少个字符串），然后输入该数量的字符串。
示例：
2
ASSSDDDAAA
AAALLLOOOLLL
然后程序应计算字符串中重复字符的数量，并提供后缀计数。
输出（基于上面的String示例）：
AS3D3A3
A3L3O3L3
它应该执行输入字符串的快捷方式。问题是我把两个字符串放在队列中，但是在程序运行时，我从未从第二个字符串得到输出，而且在第一个字符串A3中，我从未获得最后一个字符串，就像程序没有看到它（我写了ASSSDDDAAA和我得到AS3D3 idk为什么）。
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
Queue<String> kolejkaWyrazow = new LinkedList<String>();
String wyraz;
System.out.println("Podaj liczbe ciagow liczbowych");
int liczbaCiagowWyrazowych = scanner.nextInt();
for(int i = 0; i <= liczbaCiagowWyrazowych; i++) {
wyraz = scanner.nextLine();
kolejkaWyrazow.add(wyraz);
}
MarkerCutt(kolejkaWyrazow);
}
private static void MarkerCutt(Queue<String> kolejkaWyrazow) {
String box;
int countRepeats;
for(int i = 0; i <= kolejkaWyrazow.size(); i++) {
box = kolejkaWyrazow.remove();
countRepeats = 1;
for(int k = 0; k < box.length(); k++) {
if (box.charAt(k) == box.charAt(k + 1)) {
countRepeats++;
} else {
System.out.print(box.charAt(k));
if (countRepeats <= 2) {
System.out.print(box.charAt(k));
countRepeats = 1;
}
if (countRepeats >= 3) {
System.out.print(countRepeats);
countRepeats = 1;
}
}
}
}
}
}
答案 0 :(得分：0)
基于OP的原始代码，for循环存在一个问题，即对下一个字符的检查将超出String的末尾，从而导致运行时错误。
对于给定的String box中最后一个条目缺少输出，是因为在for本身的box.length()循环中生成了输出。因此，当末尾有一个重复的字符时，它永远不会显示。我怀疑可以用AAABBBC之类的东西显示相同的问题（缺少最后一个字符），该问题应显示A3B3C，但不会显示最后一个C。
稍作修改的算法将是始终在字符更改时显示该字符，然后在需要时添加计数。
private static void MarkerCutt(Queue<String> kolejkaWyrazow)
{
// the string we are processing from the queue
String box;
// number of repeats for a character in the string
int countRepeats;
while (!kolejkaWyrazow.isEmpty()) {
box = kolejkaWyrazow.remove();
countRepeats = 0;
// we hold the current character and the previous character;
// at entry, set the prevChar to something that cannot be matched
char curChar, prevChar = '\0';
// loop over the whole string; as we are not looking ahead, we
// use the full length of the String
for (int k = 0; k < box.length(); k++) {
// get the character
curChar = box.charAt(k);
// if we are not the same as the previous, then we
// may need to output the count; but output the character
// we just found, regardless
if (curChar != prevChar) {
outputCount(countRepeats);
countRepeats = 0;
System.out.print(curChar);
}
else {
++countRepeats;
}
prevChar = curChar;
}
// have to output the end of the string
outputCount(countRepeats);
System.out.println();
}
}
private static void outputCount(int countRepeats)
{
if (countRepeats > 0) {
System.out.print(countRepeats + 1);
}
}
基于显示的输入（加上AAABBBC和'AAABBBCC`：
AS3D3A3
A3L3O3L3
A3B3C
A3B3C2
在更广泛的主题上，我会考虑从方法MarkerCutt中删除实际的输出，而是使用一个返回String的方法。将输出与逻辑分离通常是一个好主意。所以：
private static void MarkerCutt(Queue<String> kolejkaWyrazow)
{
while (!kolejkaWyrazow.isEmpty()) {
String box = kolejkaWyrazow.remove();
String repString = changeToRepeatCount(box);
System.out.println(repString);
}
}
private static String changeToRepeatCount(String box)
{
// number of repeats for a character in the string
int countRepeats = 0;
// We build into this structure
StringBuilder sb = new StringBuilder();
// we hold the current character and the previous character;
// at entry, set the prevChar to something that cannot be matched
char curChar, prevChar = '\0';
// loop over the whole string; as we are not looking ahead, we
// use the full length of the String
for (int k = 0; k < box.length(); k++) {
// get the character
curChar = box.charAt(k);
// if we are not the same as the previous, then we may need
if (curChar != prevChar) {
outputCount(countRepeats, sb);
countRepeats = 0;
sb.append(curChar);
}
else {
++countRepeats;
}
prevChar = curChar;
}
// have to output the end of the string
outputCount(countRepeats, sb);
return sb.toString();
}
private static void outputCount(int countRepeats, StringBuilder sb)
{
if (countRepeats > 0) {
sb.append(countRepeats + 1);
}
}
后一种方法可以进行一些更好的/自动化的测试，因为然后可以编写执行以下操作的驱动程序：
String inp = "ASSSDDDAAA";
String exp = "AS3D3A3";
String res = changeToRepeatCount(inp);
if (! exp.equals(res)) {
System.err.println("Test Failed!");
}
（或者最好还是使用JUnit之类的东西）。这将使开发变得比每次都要键入一个值来测试代码都容易得多。只是想一想。
答案 1 :(得分：0)
document.addEventListener('DOMContentLoaded', function () {
var Modalelem = document.querySelector('.modal');
var instanceModal = M.Modal.init(Modalelem);
instanceModal.open();
});
输入数字并按Enter键时，会将空字符串添加到kolejkaWyrazow，因此我添加了“ if（wyraz.trim（）。length（）> 0）”以避免这种情况
然后我重写您的MarkerCutt方法，请参见下面的
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
Queue<String> kolejkaWyrazow = new LinkedList<String>();
String wyraz;
System.out.println("Podaj liczbe ciagow liczbowych");
int iCount = scanner.nextInt();
for (int i = 0; i <= iCount; i++) {
wyraz = scanner.nextLine();
if (wyraz.trim().length() > 0) {
kolejkaWyrazow.add(wyraz);
}
}
MarkerCutt(kolejkaWyrazow);
}
测试：
private static void MarkerCutt(Queue<String> queue) {
for (int i = 0; i <= queue.size(); i++) {
StringBuilder sb = new StringBuilder();
String box = queue.remove();
char c = box.charAt(0);
int iCount = 1;
for (int j = 1; j < box.length(); j++) {
if (box.charAt(j) == c) {
iCount++;
} else {
sb.append(c + "" + iCount);
c = box.charAt(j);
iCount = 1;
}
}
sb.append(c + "" + iCount);
System.out.println(sb.toString());
}
}
结果
Podaj liczbe ciagow liczbowych
2
ASSSDDDAAA
AAALLLOOOLLL
如果您不想打印不重复的char的计数（count = 1）：
更改
A1S3D3A3
A3L3O3L3
到
sb.append(c + "" + iCount);
测试
sb.append(c + "" + (iCount > 1 : iCount : ""));
结果：
Podaj liczbe ciagow liczbowych
2
ASSSDDDAAA
AAALLLOOOLLL