2019年全国高校程序设计挑战赛JAVA编程题5

2019年全国高校程序设计挑战赛JAVA编程题5

编程描述

给定4只包含0-9、 +、 * 的合法数学表达式(长度<1000),规定加’的优先级高于乘号 ,请输出计算结果。

输入格式

合法的数学表达式

输出格式

输出表达式的计算结果

输入样例

12*3+12*2 

输出样例

360

算法实现

Java(jdk1.7)

代码

import java.util.Scanner;
import java.util.Stack;

public class Main {
	public static Scanner sc = new Scanner(System.in);

	public static void main(String[] args) {
		String input = sc.nextLine();
		sc.close();

		double result1 = compute(input);
		System.out.println((int) result1);
	}

	public static int priority(char s) {
		switch (s) {
		case '+':
			return 2;
		case '*':
			return 1;
		default:
			return -1;
		}
	}

	public static double compute(double num1, double num2, char s) {
		switch (s) {
		case '+':
			return num1 + num2;
		case '*':
			return num1 * num2;
		default:
			return 0;

		}
	}

	public static double compute(String str) {
		double num[] = new double[20];
		int flag = 0, begin = 0, end = 0, now;
		now = -1;
		Stack<Character> st = new Stack<Character>();
		for (int i = 0; i < str.length(); i++) {
			char s = str.charAt(i);
			if (s == ' ') {

			} else if (s == '+' || s == '*') {
				if (flag == 1) {
					now += 1;
					if (end < begin) {
						num[now] = Integer.valueOf(str.substring(begin, begin + 1));
					} else {
						num[now] = Integer.valueOf(str.substring(begin, end + 1));
					}
					flag = 0;
				}
				if (s == '-') {
					if (i == 0) {
						flag = 1;
						begin = 0;
					} else if (str.charAt(i - 1) == '(' || str.charAt(i - 1) == '*' || str.charAt(i - 1) == '/') {
						flag = 1;
						begin = i;
					} else {
						if (st.empty()) {
							st.push(s);
						} else if (s == ')') {
							num[now - 1] = compute(num[now - 1], num[now], st.pop());
							now -= 1;
							st.pop();
						} else if (s == '(') {
							st.push(s);
						} else if (priority(s) <= priority(st.peek())) {
							num[now - 1] = compute(num[now - 1], num[now], st.pop());
							now -= 1;
							st.push(s);
						} else {
							st.push(s);
						}
					}
				} else if (st.empty()) {
					st.push(s);
				} else if (s == ')') {
					num[now - 1] = compute(num[now - 1], num[now], st.pop());
					now -= 1;
					st.pop();
				} else if (s == '(') {
					st.push(s);
				} else if (priority(s) <= priority(st.peek())) {
					num[now - 1] = compute(num[now - 1], num[now], st.pop());
					now -= 1;
					st.push(s);
				} else {
					st.push(s);
				}

			} else if (flag == 0) {
				flag = 1;
				begin = i;
			} else {
				end = i;
			}

		}
		if (flag == 1) {
			now += 1;
			if (end < begin) {
				num[now] = Integer.valueOf(str.substring(begin, begin + 1));
			} else {
				num[now] = Integer.valueOf(str.substring(begin, end + 1));
			}
		}
		while (now > 0) {
			num[now - 1] = compute(num[now - 1], num[now], st.pop());
			now -= 1;
		}
		return num[0];
	}
}