java:刽子手游戏重复字母
我有一个hangman项目的逻辑问题,它从用户那里获取一封信,然后搜索该信是否包含在秘密单词中。问题在于我编程的方式,如果用户在秘密单词中猜到的字母出现过多次。它将只是通过并表示它们所有。这不是我想要的,我只希望它一次更新一个正确猜测的字母的状态 我尝试了一些不同的方法,比如在java:刽子手游戏重复字母,java,string,stringbuffer,chars,Java,String,Stringbuffer,Chars,我有一个hangman项目的逻辑问题,它从用户那里获取一封信,然后搜索该信是否包含在秘密单词中。问题在于我编程的方式,如果用户在秘密单词中猜到的字母出现过多次。它将只是通过并表示它们所有。这不是我想要的,我只希望它一次更新一个正确猜测的字母的状态 我尝试了一些不同的方法,比如在状态(猜测,之后设置一个中断,但是迭代器只会转到第一个字母匹配的地方,并在那里停止 有什么简单的解决办法吗 private void compare(String str) { guessCh = str.char
状态(猜测,
之后设置一个中断,但是迭代器只会转到第一个字母匹配的地方,并在那里停止
有什么简单的解决办法吗
private void compare(String str)
{
guessCh = str.charAt(0);
char secretCh = '0';
for (int i = i2; i < secretWord.length(); i++) // Cuts the secret word into individual chars to process.
{
secretCh = secretWord.charAt(i);
// Compare the two strings.
if (guessCh == secretCh)
{
status(guessCh, i); // Sends the letter & placement to status().
}
}
}
private void比较(String str)
{
猜测=str.charAt(0);
char secretCh='0';
for(int i=i2;i
n
私有字符串状态(字符猜测、整数放置)
{
/*更新并返回状态*/
如果(猜测>='A'&&guessCh我猜这是某种作业-但无论如何,为什么不使用我猜这是某种作业-但无论如何,为什么不使用您可以使用比较方法中的状态变量来测试先前的解决方案
if (guessCh == secretCh && status.charAt(i) != secretCh)
{
status(guessCh, i);
break;
}
您可以使用比较方法中的状态变量来测试先前的解决方案
if (guessCh == secretCh && status.charAt(i) != secretCh)
{
status(guessCh, i);
break;
}
据我所知(并理解),您面临的基本问题是由compare方法中的for
循环引起的
if (guessCh == secretCh && status.charAt(i) != secretCh)
{
status(guessCh, i);
break;
}
(注意,我的示例区分大小写,您需要考虑)
我可以推荐两种基本方法
第一个是,用一次检查匹配所有出现的情况
private char guessCh;
private String secretWord;
private String status;
private String secretBuffer;
public TestStringCompare() {
secretWord = "This is a test";
// This is a copy of the secret word, this ensures that
// we always have a copy of the original.
secretBuffer = secretWord;
status = "______________";
guessCh = 'i';
compare("i");
}
private void compare(String str) {
while (secretBuffer.contains(str)) {
int foundAt = secretBuffer.indexOf(str);
status(str.charAt(0), foundAt);
// We want to remove the "guess" from our check string
// so it doesn't cause a false positive in the future
StringBuilder sb = new StringBuilder(secretBuffer);
sb.replace(foundAt, foundAt + 1, "_");
secretBuffer = sb.toString();
System.out.println(secretBuffer);
}
}
private String status(char guessCh, int placement) {
/* Update and return status. */
if (Character.isLetter(guessCh)) {
status = new StringBuffer(status).deleteCharAt(placement).toString();
status = new StringBuffer(status).insert(placement, guessCh).toString();
System.out.println("That guess is correct.");
System.out.println(status);
}
return status;
}
这将产生:
That guess is correct.
__i___________
Th_s is a test
That guess is correct.
__i__i________
Th_s _s a test
或者,替换第一次出现的猜测(据我所知,这就是你所追求的)
这将产生这个
That guess is correct.
__i___________
Th_s is a test
据我所知(并理解),您面临的基本问题是由compare方法中的for
循环引起的
if (guessCh == secretCh && status.charAt(i) != secretCh)
{
status(guessCh, i);
break;
}
(注意,我的示例区分大小写,您需要考虑)
我可以推荐两种基本方法
第一个是,用一次检查匹配所有出现的情况
private char guessCh;
private String secretWord;
private String status;
private String secretBuffer;
public TestStringCompare() {
secretWord = "This is a test";
// This is a copy of the secret word, this ensures that
// we always have a copy of the original.
secretBuffer = secretWord;
status = "______________";
guessCh = 'i';
compare("i");
}
private void compare(String str) {
while (secretBuffer.contains(str)) {
int foundAt = secretBuffer.indexOf(str);
status(str.charAt(0), foundAt);
// We want to remove the "guess" from our check string
// so it doesn't cause a false positive in the future
StringBuilder sb = new StringBuilder(secretBuffer);
sb.replace(foundAt, foundAt + 1, "_");
secretBuffer = sb.toString();
System.out.println(secretBuffer);
}
}
private String status(char guessCh, int placement) {
/* Update and return status. */
if (Character.isLetter(guessCh)) {
status = new StringBuffer(status).deleteCharAt(placement).toString();
status = new StringBuffer(status).insert(placement, guessCh).toString();
System.out.println("That guess is correct.");
System.out.println(status);
}
return status;
}
这将产生:
That guess is correct.
__i___________
Th_s is a test
That guess is correct.
__i__i________
Th_s _s a test
或者,替换第一次出现的猜测(据我所知,这就是你所追求的)
这将产生这个
That guess is correct.
__i___________
Th_s is a test
status.charAt(i)!=secretCh的测试使其跳过已解决的字母,break
使其在解决一个新字母时停止。status.charAt(i)的测试!=secretCh
使其跳过已解决的字母,而break
使其在解决一个新字母时停止。