最近在看《算法》這本書，正好看到一個計算表達式的問題，于是就打算寫一下，也正好熟悉一下Java集合框架的使用，大致測試了一下，沒啥問題。

import java.util.*; /** * 用來計算表達式* 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));* 注意點：* 2.輸入的表達書不能還有空格,括號必須匹配* 基本思想：* 1.建立操作數棧以及操作符棧* 2.先去括號，每次遇到')'時，就退棧，直到遇到'('* 3.然后處理括號中的表達式，先處理優先級高的，即*、/* 4.處理好高優先級操作符之后，就處理+、-這種操作符* 5.對以上的運算結果入棧，繼續，處理完所有的(、)之后* 6.然后再次求一般的表達式即可*/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) {//匹配非數字，將(、)、+、-、*、/作為分隔符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))) {//匹配到右括號，需要計算括號中的內容if (str.charAt(i) == ')') {Deque<Character > ops_tmp = new LinkedList<Character >();while (!ops.isEmpty() && ops.peek() != '(') {ops_tmp.offerFirst(ops.pop());}//'('退棧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) {//操作數數目=操作符數目+1int 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;}}

轉載于:https://www.cnblogs.com/wally/p/4477036.html

總結

以上是生活随笔為你收集整理的Java集合框架练习-计算表达式的值的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。