Java 从字符串中提取整数和字符
我正在使用堆栈数据结构解决数学表达式问题,我一直在从字符串中提取数字和数学符号。我想做的是制作一个程序来计算给定的表达式 我的问题是,如何将字符串中的所有数字和符号提取到新的字符串数组中Java 从字符串中提取整数和字符,java,expression,extraction,Java,Expression,Extraction,我正在使用堆栈数据结构解决数学表达式问题,我一直在从字符串中提取数字和数学符号。我想做的是制作一个程序来计算给定的表达式 我的问题是,如何将字符串中的所有数字和符号提取到新的字符串数组中 Input: 12*1*145*2+8*1*1+2*3+2+4 新字符串数组应为: String[] expArray={"12","*","1","*","145","*","2","+","8","*","1","*","1","+","2","*","3","+","2","+","4"} 我尝试了n
Input: 12*1*145*2+8*1*1+2*3+2+4
新字符串数组应为:
String[] expArray={"12","*","1","*","145","*","2","+","8","*","1","*","1","+","2","*","3","+","2","+","4"}
我尝试了nextInt()
,split()
,next()
,nextByte()
,但没有成功。别担心别的事。我刚才提到这个问题只是为了弄清楚我在说什么
更新:我不需要像
array[I]*10+array[I+1]
或类似的构建整数的解决方案。我需要的是更好的解决方案。但由于其他人可能感兴趣,我还是发布了
import java.util.*;
public class Test {
public static void main(String... args) {
String s = "12*1*145*2+8*1*1+2*3+2+4";
List<String> tokens = new ArrayList<>();
// for building numbers
StringBuilder builder = new StringBuilder();
// assume there are only 2 types (number and operators)
boolean isNumber = false;
for (char c : s.toCharArray()) {
if (Character.isDigit(c)) {
if (!isNumber) {
builder = new StringBuilder();
isNumber = true;
}
builder.append(c);
} else {
if (isNumber) {
tokens.add(builder.toString());
isNumber = false;
}
// assume operators are exactly 1 char
tokens.add(String.valueOf(c));
}
}
if (isNumber)
tokens.add(builder.toString());
System.out.println(tokens);
}
}
更通用的方法是使用类型的enum
来保持当前状态,而不是使用布尔值
,并对所有类型使用builder
,以便它们的标记不限于1个字符
然后您可能需要为其他人使用Regex构建一个…解决方案:
String inputStr = "12*1*145*2+8*1*1+2*3+2+4";
String inputPattern = "(\\d+|[+-/*]{1})";
List<String> strList=new ArrayList<>();
Pattern patternObj = Pattern.compile(inputPattern);
Matcher matcherObj = patternObj.matcher(inputStr);
while (matcherObj.find()) {
strList.add(matcherObj.group(0));
}
System.out.println(strList);
String inputStr=“12*1*145*2+8*1*1+2*3+2+4”;
String inputPattern=“(\\d+|[+-/*]{1})”;
List strList=new ArrayList();
Pattern patternObj=Pattern.compile(inputPattern);
Matcher matcherObj=patternObj.Matcher(inputStr);
而(matcherObj.find()){
strList.add(matcherObj.group(0));
}
System.out.println(strList);
请保留@MichaelLihs的编辑,它非常好。已向版主标记两次积极的回滚。非常感谢先生。这对我帮助很大。
String inputStr = "12*1*145*2+8*1*1+2*3+2+4";
String inputPattern = "(\\d+|[+-/*]{1})";
List<String> strList=new ArrayList<>();
Pattern patternObj = Pattern.compile(inputPattern);
Matcher matcherObj = patternObj.matcher(inputStr);
while (matcherObj.find()) {
strList.add(matcherObj.group(0));
}
System.out.println(strList);