题解 | 四则运算
四则运算
https://www.nowcoder.com/practice/9999764a61484d819056f807d2a91f1e
import java.util.*;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
// 注意 hasNext 和 hasNextLine 的区别
while (in.hasNext()) { // 注意 while 处理多个 case
String s = in.nextLine();
int n = s.length();
int num1 = 0;
int o1 = 1;
int num2 = 1;
int o2 = 1;
Stack<Integer> stk = new Stack<>();
for (int i = 0; i < n; i++) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
int cur = 0;
while (i < n && Character.isDigit(s.charAt(i))) {
cur = cur * 10 + s.charAt(i) - '0';
i++;
}
i--; // 定位回当前字母
num2 = o2 == 1 ? num2 * cur : num2 / cur;
} else if (c == '*' || c == '/') {
o2 = c == '*' ? 1 : -1;
} else if (c == '(' || c == '{' || c == '[') {
stk.push(num1);
stk.push(o1);
stk.push(num2);
stk.push(o2);
// 重置
num1 = 0;
o1 = 1;
num2 = 1;
o2 = 1;
} else if (c == '+' ||
c == '-') { //遇到加减,说明可以开始计算,计算num1并对定义其他几个变量
if (c == '-' && (i == 0 || s.charAt(i - 1) == '(' || s.charAt(i - 1) == '[' ||
s.charAt(i - 1) == '{')) {
o1 = -1;
continue;
}
num1 = num1 + o1 * num2;
o1 = (c == '+' ? 1 : -1);
num2 = 1;
o2 = 1;
}else{ // 遇到右括号了
int cur = num1+o1*num2;
o2 = stk.pop();
num2 = stk.pop();
o1 = stk.pop();
num1 = stk.pop();
num2 = o2==1?num2*cur:num2/cur;
}
}
System.out.println(num1+o1*num2);
}
}
}
常见操作,遇到右括号就把之前的计算结果入栈,重新计算里面;出来后继续计算;

