我在完成这项任务时遇到了麻烦：
给定一个字符串，将第一个出现的'a'替换为“x”，将第二个出现的'a'替换为“xx”，将第三个出现的'a'替换为“xxx”。在第三次出现之后，再次用“x”，“xx”，“xxx”......等开始替换模式;但是，如果一个'a'后跟一行中超过2个其他'a'字符，那么在'a'之后不要再替换'a'字符。
不允许使用替换方法。
aTo123X（“ababba”）→“xbxxbbxxx”
aTo123X（“anaceeacdabnanbag”）→“xnxxceexxxcdxbnxxnbxxxg”
aTo123X（“aabaaaavfaajaaj”）→“xxxbxxxaaavfaajaaj”
aTo123X（“pakaaajaaaamnbaa”）→“pxkxxxxxxjxxaaamnbaa”
aTo123X（“aaaak”）→“xaaak”
我的代码的输出是包含的，x是添加但不是正确的x的数量。
public String aTo123X(String str) {
/*
Strategy:
get string length of the code, and create a for loop in order to find each individual part of the String chars.check for a values in string and take in pos of the a.
if one of the characters is a
replace with 1 x, however, there aren't more than 2 a's immediately following first a and as it keeps searching through the index, add more x's to the original string, but set x value back to 1 when x reaches 3.
if one of characters isn't a,
leave as is and continue string.
*/
String xVal = "";
String x = "x";
String output = "";
for (int i = 0; i < str.length(); i++){
if( str.charAt(i) == 'a'){
output += x;
str.substring(i+1, str.length());
}
output += str.charAt(i);
}
return output;
}
答案 0 :(得分：3)
这是执行相同操作的代码。我已经对代码进行了评论，以解释它的作用
public class ReplaceChar {
public static void main(String... args){
String[] input =new String[]{"ababba","anaceeacdabnanbag","aabaaaavfaajaaj"};
StringBuilder result = new StringBuilder();
for (int i= 0; i < input.length;i++){
result.append(getReplacedA(input[i]));
result.append("\n");
}
System.out.println(result);
}
private static String getReplacedA(String withA){
// stringBuilder for result
StringBuilder replacedString = new StringBuilder();
// counting the number of time char 'a' occurred in String for replacement before row of 'aaa'
int charACount = 0;
// get the first index at which more than two 'aa' occurred in a row
int firstIndexOfAAA = withA.indexOf("aaa") + 1;
// if 'aaa' not occurred no need to add the rest substring
boolean addSubRequired = false;
// if the index is 0 continue till end
if (firstIndexOfAAA == 0)
firstIndexOfAAA = withA.length();
else
addSubRequired = true;
char[] charString = withA.toCharArray();
//Replace character String[] array
String[] replace = new String[]{"x","xx","xxx"};
for(int i = 0; i < firstIndexOfAAA; i++){
if (charString[i] == 'a'){
charACount++;
charACount = charACount > 3 ? 1 : charACount ;
// add the number x based on charCount
replacedString.append(replace[charACount - 1]);
}else{
replacedString.append(charString[i]);
}
}
// if the String 'aaa' has been found previously add the remaining subString
// after that index
if (addSubRequired)
replacedString.append(withA.substring(firstIndexOfAAA));
// return the result
return replacedString.toString();
}
}
输出：
xbxxbbxxx
xnxxceexxxcdxbnxxnbxxxg
xxxbxxxaaavfaajaaj
编辑：一些改进你可以在getReplacedA（）函数中为一些极端情况做出：
检查char＆＃39; a＆＃39;是否存在于String中，如果不是只返回字符串则无需再做任何事情。
使用IgnoreCase可以避免大写或小写的可能性。
答案 1 :(得分：1)
首先，字符串是不可变的，因此以下语句不执行任何操作
str.substring(i+1, str.length());
我想你想做：
str = str.substring(i+1, str.length());
然而，即使在修复之后，您的程序仍然无效。我无法理解你的解决方案。 1）你连续检测不到3个以上。 2）你没有附加“xx”或“xxx”
这是我的版本，到目前为止对我有用：
public static void main(String[] args) {
System.out.println(aTo123X("ababba")); // "xbxxbbxxx"
System.out.println(aTo123X("anaceeacdabnanbag")); // "xnxxceexxxcdxbnxxnbxxxg"
System.out.println(aTo123X("aabaaaavfaajaaj")); // "xxxbxxxaaavfaajaaj"
}
public static String aTo123X(String str) {
String output = "";
int aOccurrence = 0;
String[] xs = {"x", "xx", "xxx"};
for (int i = 0; i < str.length(); ++i) {
if (str.charAt(i) == 'a') {
output += xs[aOccurrence % 3]; // append the x's depending on the number of a's we have seen, modulus 3 so that it forms a cycle of 3
if (i < str.length() - 3 && str.charAt(i + 1) == 'a' && str.charAt(i + 2) == 'a' && str.charAt(i + 3) == 'a') {//if an 'a' is followed by more than 2 other 'a' characters in a row
output += str.substring(i + 1);
break;
} else {
++aOccurrence; // increment the a's we have encountered so far
}
} else {
output += str.charAt(i); // append the character if it is not a
}
}
return output;
}
答案 2 :(得分：1)
public class NewClass {
public static void main(String[] args) {
System.out.println(aTo123X("ababba")); // "xbxxbbxxx"
System.out.println(aTo123X("anaceeacdabnanbag")); // "xnxxceexxxcdxbnxxnbxxxg"
System.out.println(aTo123X("aabaaaavfaajaaj")); //xxxbxxxaaavfaajaaj
}
public static String aTo123X(String str) {
String output = "";
int aCount = 0;
int inRow = 0;
for (int i = 0; i < str.length();) {
if (str.charAt(i) == 'a') {
if (inRow <= 1) {
inRow++;
aCount++;
if (aCount == 1) {
output += "x";
} else if (aCount == 2) {
output += "xx";
} else {
output += "xxx";
aCount = 0;
}
boolean multiple = ((i + 1) < str.length()) && (str.charAt(i + 1) == 'a')
&& ((i + 2) < str.length()) && (str.charAt(i + 2) == 'a');
if (multiple) {
i++;
while (i < str.length()) {
output += str.charAt(i++);
}
return output;
}
} else {
output += str.charAt(i);
}
} else {
output += str.charAt(i);
inRow = 0;
}
i++;
}
return output;
}
}
答案 3 :(得分：1)
我在代码中以注释的形式指出了代码中的问题。
public String aTo123X(String str) {
//You are not using xVal variable in your code, hence it's obsolete
String xVal = "";
//You don't need x variable as you can simply use string concatenation
String x = "x";
String output = "";
for (int i = 0; i < str.length(); i++) {
/**
* Here, in "if" block you have not implmented any logic to replace the 2nd and
* 3rd occurence of 'a' with 'xx' and 'xxx' respectively. Also, substring() returns
* the sub-string of a string but you are not accepting that string anywhere, and
* you need not even use sub-string as "for" loop will cycle through all the
* characters in the string. If use sub-string method you code will only process
* alternative characters.
*/
if( str.charAt(i) == 'a') {
output += x;
str.substring(i+1, str.length());
}
/**
* Because of this statement a's are also returned, because this statement gets
* in both scenarios, whether the current character of string is a or not.
* But, this statement should get executed only when current character of the
* string is 'a'. So, in terms of coding this statement gets executed no matter
* "if" loop is executed or not, but it should get executed only when "if" loop
* is not executed. So, place this statement in else block.
*/
output += str.charAt(i);
}
return output;
}
我已经为你实现了逻辑。这是您的问题的解决方案，只需复制并运行它。它通过了所有指定的测试用例。
public String aTo123X(String str) {
String output = "";
int count = 1;
boolean flag = true;
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == 'a' && flag == true) {
switch(count) {
case 1: output += "x";
count++;
break;
case 2: output += "xx";
count++;
break;
case 3: output += "xxx";
count = 1;
break;
}
if ((str.charAt(i+1) == 'a' && str.charAt(i+2) == 'a') == true) {
flag = false;
}
}
else {
output += str.charAt(i);
}
}
return output;
}
答案 4 :(得分：0)
我已经编辑了我的答案。这个是给出正确的解决方案：
public static void main (String[] args) throws InterruptedException, IOException, JSONException {
System.out.println(aTo123X("ababba")); //xbxxbbxxx
System.out.println(aTo123X("anaceeacdabnanbag")); //xnxxceexxxcdxbnxxnbxxxg
System.out.println(aTo123X("aabaaaavfaajaaj")); //xxxbxxxaaavfaajaaj
}
public static String aTo123X(String str) {
String x = "x";
String xx = "xx";
String xxx = "xxx";
int a = 1;
int brek = 0;
String output = "";
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == 'a' && a == 1) {
output += x;
str.substring(i+1, str.length());
a = 2;
try {
if(str.charAt(i+1) == 'a' && str.charAt(i+2) == 'a')
brek += 1;
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
else if(str.charAt(i) == 'a' && a == 2) {
output += xx;
str.substring(i+1, str.length());
a = 3;
try {
if(str.charAt(i+1) == 'a' && str.charAt(i+2) == 'a')
brek += 1;
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
else if(str.charAt(i) == 'a' && a == 3) {
output += xxx;
str.substring(i+1, str.length());
a = 1;
try {
if(str.charAt(i+1) == 'a' && str.charAt(i+2) == 'a')
brek += 1;
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
else {
output += str.charAt(i);
brek = 0;
}
if(brek>0) {
output += str.substring(i+1);
break;
}
}
return output;
}
答案 5 :(得分：0)
我使用地图存储替换位置
public static void main(String[] args) {
System.out.println(aTo123X("ababba"));//xbxxbbxxx
System.out.println(aTo123X("anaceeacdabnanbag"));//xnxxceexxxcdxbnxxnbxxxg
System.out.println(aTo123X("aabaaaavfaajaaj"));//xxxbxxxaaavfaajaaj
}
public static String aTo123X(String str){
String res = "";
int nthReplace = 1; //Integer to store the nth occurence to replace
//Map to store [key == position of 'a' to replace]
//[value == x or xx or xxx]
Map<Integer, String> toReplacePos = new HashMap<>();
//The loop to know which 'a' to replace
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == 'a'){
toReplacePos.put(i, nthReplace % 3 == 1 ? "x": (nthReplace % 3 == 2 ? "xx": "xxx"));
nthReplace++;
//Break if an 'a' is followed by more than 2 other 'a'
try {
if((str.charAt(i+1) == 'a')
&& (str.charAt(i+2) == 'a')
&& (str.charAt(i+3) == 'a')){
break;
}
} catch (StringIndexOutOfBoundsException e) {
}
}
}
//Do the replace
for (int i = 0; i < str.length(); i++) {
res += toReplacePos.containsKey(i) ? toReplacePos.get(i) : str.charAt(i);
}
return res;
}