如何解决检查字符是整数还是字母
| 我正在使用Java修改文件。这是我要完成的工作： 如果在读取时检测到＆符号以及整数，我想删除＆符号并将整数转换为二进制。 如果在读取时检测到＆符号以及一个（随机）单词，我想删除＆符号并用整数16替换该单词，并且是否要与＆符号一起使用不同的字符串，我想将数字1设置为大于整数16。 这是我的意思的一个例子。如果输入的文件包含以下字符串：&myword
&4
&anotherword
&9
&yetanotherword
&10
&myword
&0000000000010000 (which is 16 in decimal)
&0000000000000100 (or the number \'4\' in decimal)
&0000000000010001 (which is 17 in decimal,since 16 is already used,so 16+1=17)
&0000000000000101 (or the number \'9\' in decimal)
&0000000000010001 (which is 18 in decimal,or 17+1=18)
&0000000000000110 (or the number \'10\' in decimal)
&0000000000010000 (which is 16 because value of myword = 16)
for (i=0; i<anyLines.length; i++) {
char[] charray = anyLines[i].toCharArray();
for (int j=0; j<charray.length; j++)
if (Character.isDigit(charray[j])) {
anyLines[i] = anyLines[i].replace(\"&\",\"\");
anyLines[i] = Integer.toBinaryString(Integer.parseInt(anyLines[i]);
}
else {
continue;
}
if (Character.isLetter(charray[j])) {
anyLines[i] = anyLines[i].replace(\"&\",\"\");
for (int k=16; j<charray.length; k++) {
anyLines[i] = Integer.toBinaryString(Integer.parseInt(k);
}
}
}
}
解决方法
Character.isLetter() //tests to see if it is a letter
Character.isDigit() //tests the character to
&
indexOf(\'%\')
Map<String,Integer>
Map<String,Integer> usedWords = new HashMap<String,Integer>();
List<String> output = new ArrayList<String>();
int wordIncrementer = 16;
String[] arr = test.split(\"\\n\");
for(String s : arr)
{
if(s.startsWith(\"&\"))
{
String line = s.substring(1).trim(); //Removes &
try
{
Integer lineInt = Integer.parseInt(line);
output.add(\"&\" + Integer.toBinaryString(lineInt));
}
catch(Exception e)
{
System.out.println(\"Line was not an integer. Parsing as a String.\");
String outputString = \"&\";
if(usedWords.containsKey(line))
{
outputString += Integer.toBinaryString(usedWords.get(line));
}
else
{
outputString += Integer.toBinaryString(wordIncrementer);
usedWords.put(line,wordIncrementer++);
}
output.add(outputString);
}
}
else
{
continue; //Nothing indicating that we should parse the line.
}
}
String input = \"&myword\\n&4\\n&anotherword\\n&9\\n&yetanotherword\\n&10\\n&myword\";
String[] lines = input.split(\"\\n\");
int wordValue = 16;
// to keep track words that are already used
Map<String,Integer> wordValueMap = new HashMap<String,Integer>();
for (String line : lines) {
// if line doesn\'t begin with &,then ignore it
if (!line.startsWith(\"&\")) {
continue;
}
// remove &
line = line.substring(1);
Integer binaryValue = null;
if (line.matches(\"\\\\d+\")) {
binaryValue = Integer.parseInt(line);
}
else if (line.matches(\"\\\\w+\")) {
binaryValue = wordValueMap.get(line);
// if the map doesn\'t contain the word value,then assign and store it
if (binaryValue == null) {
binaryValue = wordValue;
wordValueMap.put(line,binaryValue);
wordValue++;
}
}
// I\'m using Commons Lang\'s StringUtils.leftPad(..) to create the zero padded string
String out = \"&\" + StringUtils.leftPad(Integer.toBinaryString(binaryValue),16,\"0\");
System.out.println(out);
&0000000000010000
&0000000000000100
&0000000000010001
&0000000000001001
&0000000000010010
&0000000000001010
&0000000000010000
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