Java集合框架练习-计算表达式的值

最近在看《算法》这本书,正好看到一个计算表达式的问题,于是就打算写一下,也正好熟悉一下Java集合框架的使用,大致测试了一下,没啥问题。

import java.util.*;
/*
 * 
 * 用来计算表达式
 * for example: 1+2*3*(4+3*1)-3*1+2+3/1;
 * (1+2*2-2*1*3*(1-1))*(1-2+3*(4+0));
 * 注意点:
 * 2.输入的表达书不能还有空格,括号必须匹配
 * 基本思想:
 * 1.建立操作数栈以及操作符栈
 * 2.先去括号,每次遇到')'时,就退栈,直到遇到'('
 * 3.然后处理括号中的表达式,先处理优先级高的,即*、/
 * 4.处理好高优先级操作符之后,就处理+、-这种操作符
 * 5.对以上的运算结果入栈,继续,处理完所有的(、)之后
 * 6.然后再次求一般的表达式即可
 */

public class CalExpression {
	
	private Stack<Double > vals = new Stack<Double >();
	private Stack<Character > ops = new Stack<Character >();
	
	public static void main(String[] args) {
		CalExpression obj = new CalExpression();
		obj.input();
	}
	
	public void input() {
		Scanner in =  new Scanner(System.in);
		while (in.hasNext()) {
			pushStack(in.next());
		}
	}
	
	public boolean check(char ch) {
		if (ch == '(' || ch == ')' || ch == '+' || ch == '-'
				|| ch == '*' || ch == '/') {
			return true;
		}
		return false;
	}
	
	
	public void pushStack(String str) {
		//匹配非数字,将(、)、+、-、*、/作为分隔符
		String[] strNum = str.split("[^0-9]"); 
	
		Queue<Double > que = new LinkedList<Double >();
		
		for (int i = 0; i < strNum.length; ++i) {
			if (!strNum[i].equals("")) {
				que.offer(Double.parseDouble(strNum[i]));
			}
		}
		
		boolean flag = false;
		for (int i = 0; i < str.length(); ++i) {
			if (check(str.charAt(i))) {
				//匹配到右括号,需要计算括号中的内容
				if (str.charAt(i) == ')') {
					Deque<Character > ops_tmp = new LinkedList<Character >();
					while (!ops.isEmpty() && ops.peek() != '(') {
						ops_tmp.offerFirst(ops.pop());
					}
					//'('退栈
					ops.pop();
					calExpress(ops_tmp);
				} else {
					ops.push(str.charAt(i));
				}
				flag = false;
				
			} else if (!flag) {
				vals.push(que.poll());
				flag = true;
			}
		}
		
		double value = getValue(vals.iterator(), ops.iterator());
		
		System.out.println(value);
		vals.clear();
		ops.clear();
		
	}
	
	
	public void calExpress(Deque<Character > deq_ops) {
		//操作数数目=操作符数目+1
		int numCount = deq_ops.size() + 1;
		
		Deque<Double > deq_num = new LinkedList<Double >();
		while (numCount > 0 && !vals.isEmpty()) {
			deq_num.offerFirst(vals.pop());
			numCount--;
		}
		
		double value = getValue(deq_num.iterator(), deq_ops.iterator());
		vals.push(value);
	}
	
	
	
	public double getValue(Iterator it_num, Iterator it_ops) {
		Deque<Double > vals = new LinkedList<Double >();
		Deque<Character > ops = new LinkedList<Character >();
		
		vals.offer((double)it_num.next());
		while (it_num.hasNext()) {
			char ch = (char)it_ops.next();
			if (ch == '+' || ch == '-') {
				vals.offer((double)it_num.next());
				ops.offer(ch);
			} else if (ch == '*' || ch == '/') {
				double num = vals.pollLast();
				if (ch == '*') {
					vals.offer(num * (double)it_num.next());
				} else {
					vals.offer(num / (double)it_num.next());
				}
			}
		}
		
		double value = vals.pollFirst();
		while (!vals.isEmpty() && !ops.isEmpty()) {
			if ((char)ops.pollFirst() == '+') {
				value += vals.pollFirst();
			} else {
				value -= vals.pollFirst();
			}
		}
		
		return value;
	}

}


posted @ 2015-01-30 21:04  ihge2k  阅读(463)  评论(0编辑  收藏  举报