题解 | #表达式求值#
表达式求值
https://www.nowcoder.com/practice/c215ba61c8b1443b996351df929dc4d4
import java.util.*;
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
* 返回表达式的值
* @param s string字符串 待计算的表达式
* @return int整型
*/
public static int solve(String s) {
//首先定义两个栈
//操作数栈
LinkedList<Integer> numStack = new LinkedList<>();
LinkedList<Character> opsStack = new LinkedList<>();
int len = s.length();
for (int i = 0; i < len; i++) {
char ch = s.charAt(i);
if (ch == ' ') {
continue;
}
//先判断一下是不是数字
if (isNumber(ch)) {
//如果是数字,则需要遍历一下,看周围的数字是否遍历完成,例如26
int num = 0;
while (i < len && isNumber(s.charAt(i))) {
num = num * 10 + s.charAt(i) - '0';
i++;
}
numStack.push(num);
//因为下一轮for循环自己会i++,所以这里要减去
i--;
} else if (ch == '(') {
//遇到左括号,直接入栈就好了
opsStack.push(ch);
} else if (ch == ')') {
//遇到右括号了,如果遇不到左括号就一直弹出来
while (!opsStack.isEmpty() && opsStack.peek() != '(') {
Character op = opsStack.pop();
Integer a = numStack.pop();
Integer b = numStack.pop();
int r = res(op, b, a);
numStack.push(r);
}
//弹出左括号
opsStack.pop();
} else {
//如果扫描的操作符的优先级不大于栈顶的操作符,那么就弹栈
while (!opsStack.isEmpty() && priority(ch) <= priority(opsStack.peek())) {
Character op = opsStack.pop();
Integer a = numStack.pop();
Integer b = numStack.pop();
int r = res(op, b, a);
numStack.push(r);
}
//把操作符入栈
opsStack.push(ch);
}
}
//把剩余的弹出
while (!opsStack.isEmpty()) {
Character op = opsStack.pop();
Integer a = numStack.pop();
Integer b = numStack.pop();
int r = res(op, b, a);
numStack.push(r);
}
return numStack.pop();
}
//定义一下操作符和数字的运算结果
public static int res(char c, int o1, int o2) {
if (c == '+') {
return o1 + o2;
}
if (c == '-') {
return o1 - o2;
}
if (c == '/') {
return o1 / o2;
}
if (c == '*') {
return o1 * o2;
}
return 0;
}
//判断操作数的优先级
public static int priority(char c) {
if (c == '+' || c == '-') {
return 1;
}
if (c == '*' || c == '/') {
return 2;
}
return 0;
}
//首先写一个判断是不是数字
public static boolean isNumber(char c) {
return c >= '0' && c <= '9';
}
}

