日期:2014-05-20 浏览次数:20581 次
String str = "2.5*4+(2*5)"; Pattern p; Matcher m; p = Pattern.compile("\\+|\\-|\\*|\\/|\\(|\\)"); m = p.matcher(str); while(m.find()){ System.out.println(m.group()); } p = Pattern.compile("\\d+(\\.\\d+)*"); m = p.matcher(str); while(m.find()){ System.out.println(m.group()); }
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其实不需要拆分数组,一个一个字符解析比较好
因为拆分数组的话,可能会有问题,比如
1**3 8
拆分成数组来做,程序可能会错误地解析成
1*3*8
------解决方案--------------------
楼主可以去看看我的这篇博客,里面有完整的代码,你可以参考一下
http://blog.csdn.net/antineutrino/article/details/6763722
------解决方案--------------------
中缀表达式转后缀表达式
分割字符看你想怎么解析,也就是说 2.5合法,2. 5算不算合法的问题,如果中间的存在空格可以合法,那么去掉所有的空格后,在把字符串变成字符数组,然后遍历该字符数组即可,如果中间存在空格算不合法,那么做法一样,只是在遍历的过程中要自己判断空格符
给段小例子
import java.util.*; public class Test { static final String symbol = "+-*/()"; //运算符 static final String[] priority = {"+-", "*/", "()"}; //运算符优先级 static Comparator<String> comp = new Comparator<String>() { //运算符比较器 public int compare(String s1, String s2) { int n1=0, n2=0; for (int i=0; i<priority.length; i++) { if (priority[i].indexOf(s1) >= 0) {n1 = i;} if (priority[i].indexOf(s2) >= 0) {n2 = i;} } return (n1 - n2); } }; public static void main(String[] args) throws Throwable { String exp = "2.5*4+(2*5)"; List<String> list = analyze(exp); //中缀转后缀 System.out.println(list); double result = cacl(list); //计算结果 System.out.printf("%.2f\n", result); } public static List<String> analyze(String exp) throws Exception { //分析 if (exp == null) { throw new Exception ("illegal parameter."); } exp = exp.replaceAll("\\s*", ""); //去掉所有的空格(为了方便中间存在空格算合法) List<String> list = new ArrayList<String>(); Stack<String> sym = new Stack<String>(); StringBuilder buf = new StringBuilder(); for (char c : exp.toCharArray()) { if (symbol.indexOf(c) >= 0) { //如果是运算符 if (buf.length() > 0) { //如果有操作数 String v = buf.toString(); if (! v.matches("\\d+([.]\\d+)?")) { throw new Exception ("illegal varaible("+v+")."); } list.add(v); buf.delete(0, buf.length()); } if (c == '(') { sym.push(String.valueOf(c)); } else if (c == ')') { String last = ""; while (sym.size() > 0) { last = sym.pop(); if (last.equals("(")) { break; } else { list.add(last); } } if (!"(".equals(last)) { throw new Exception ("illigal express."); } } else if (sym.size() > 0) { String s = String.valueOf(c); String last = sym.peek(); if (last.equals("(") || comp.compare(s, last) > 0) { sym.push(s); } else { last = sym.pop(); list.add(last); sym.push(s); } } else { sym.push(String.valueOf(c)); } } else { //不是运算符则当作操作数(因为已经去除所有空格,这里不再需要判断空格) buf.append(c); } } if (buf.length() > 0) { list.add(buf.toString()); } while (sym.size() > 0) { String last = sym.pop(); if ("()".indexOf(last) >= 0) { throw new Exception ("illigal express."); } list.add(last); } return list; } public static double cacl(List<String> list) throws Exception { //计算 Stack<Double> val = new Stack<Double>(); double result = 0; while (list.size() > 0) { String s = list.remove(0); if (symbol.indexOf(s) >= 0) { double d1 = val.pop(); double d2 = val.pop(); if ("+".equals(s)) { result = d2 + d1; } else if ("-".equals(s)) { result = d2 - d1; } else if ("*".equals(s)) { result = d2 * d1; } else if ("/".equals(s)) { result = d2 / d1; } else { throw new Exception ("illigal symbol("+s+")."); } val.push(result); } else { if (!s.matches("\\d+([.]\\d+)?")) { throw new Exception ("illigal variable("+s+")."); } val.push(Double.valueOf(s)); } } return result; } }