表达式求值

A=input().replace('"','') print(eval(A))

递归

from collections import deque


class Solution:
    def solve(self, s):
        s = deque(s)

        def dfs():
            num, sign, stk = 0, '+', []
            while s:
                ch = s.popleft()
                if ch == '(':
                    num = dfs()

                if ch.isdigit():
                    num = num * 10 + int(ch)

                if (not ch.isdigit() and ch != ' ') or not s:
                    if sign == '+':
                        stk.append(num)
                    elif sign == '-':
                        stk.append(-num)
                    elif sign == '*':
                        stk.append(stk.pop() * num)
                    num, sign = 0, ch

                if ch == ')':
                    break

            return sum(stk)

        return dfs()

// 第一种方法
function solve( s ){
    let stack = [],i = 0,sign='+',num=0; 
    while(i < s.length){
        if(s[i]=='('){
            let flag= 1, start = i+1;
            while(flag>0){
                i++;
                if(s[i]==')') flag--;
                if(s[i]=='(') flag++;
            }
            num = solve(s.slice(start,i));
            i--;
        }else if(s[i]>='0'&&s[i]<='9'){
            num = num*10 + Number(s[i]);
        }
        if((s[i]<'0'|| s[i]>'9') || i==s.length-1){
            if(sign=='*') stack.push(Number(stack.pop())*num);
            if(sign=='+') stack.push(Number(num));
            if(sign=='-') stack.push(Number(num)*-1);
            if(sign=='/') stack.push(Number(stack.pop())/num);
            sign = s[i];
            num = 0;
        }
        i++;
    }
    return stack.reduce((total,n)=>{return total+n},0);
}
// 第二种
function solve ( s ) {
    return eval(s);
}
// 第三种
function solve ( s ) {
    let func = new Function(`return ${s}`);
    return func();
}

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 int solve(String s)
    {

        // 请写一个整数计算器，支持加减乘三种运算和括号。
        // write code here
        //idea the () could be regarded as a computing element using the recursion method
        Stack<Integer> stack = new Stack<>();
        int number = 0;
        int sum = 0;
        char sign = '+';
        char[] c = s.toCharArray();
        int n = s.length();
        for (int i = 0; i < n; i++)
        {
            char ele = c[i];
            //process the numerical situation
            if (Character.isDigit(ele))
            {
                number = number * 10 + ele - '0';
            }
            //process the () situation
            if (ele == '(')
            {
                int j = i + 1;
                int counterPar = 1;
                String subPar = "";
                //extract the most outer group and recursevely preocess
                while (counterPar > 0)
                {
                    if (c[j] == '(')
                    {
                        counterPar++;
                    }
                    if (c[j] == ')')
                    {
                        counterPar--;
                    }
                    j++;
                }
                subPar = s.substring(i + 1, j);
                number=solve(subPar);
                i = j-1;
            }
            //real work block
            if (ele != ' ' && !Character.isDigit(ele) || i == n - 1)
            {
                if (sign == '+')
                {
                    stack.push(number);
                }
                else if (sign == '-')
                {
                    stack.push(-1 * number);
                }
                else if (sign == '*')
                {
                    stack.push(stack.pop() * number);
                }
                else if (sign == '/')
                {
                    stack.push(stack.pop() / number);
                }
                //change the sign and number
                number = 0;
                sign = ele;
            }

        }
        while (!stack.isEmpty())
        {
            sum+=stack.pop();
        }
        return sum;
    }
}

HashMap<Character ,Integer> map=new HashMap<Character ,Integer>(){{
    put('+' ,1);
    put('-' ,1);
    put('*' ,2);
}};

public int solve (String s) {
    // write code here
    Stack<Character> s1=new Stack<>();
    Stack<Integer> s2=new Stack<>();
    s2.push(0);
    char[] cs=s.replaceAll(",","").toCharArray();

    for(int i=0;i<cs.length;i++){
        // 1、遇到左括号
        if(cs[i]=='('){
            s1.push('(');
        }else if(Character.isDigit(cs[i])){ //2、遇到数字
            int num=0;
            while(i<cs.length && Character.isDigit(cs[i])){
                num=num*10+(cs[i++]-'0');
            }
            i--;
            s2.push(num);
        }else if(cs[i]==')'){// 3、遇到右括号
            while(s1.peek()!='('){
                cacl(s1 ,s2);
            }
            s1.pop();
        }else{// 4、遇到计算符号
            if(i>0 && cs[i-1]=='('){ // 4.1 如果左括号后紧跟一个计算符号，需要在符号前加个0 ，便于进行计算
                s2.push(0);
            } // 4.2 符号优先级高的会先进行计算
            while(!s1.isEmpty() && s1.peek()!='(' && map.get(cs[i])<=map.get(s1.peek())){
                cacl(s1 ,s2);
            } // 4.3 处理完紧跟左括号和优先级计算，需要将当前的符号压栈
            s1.push(cs[i]);
        }
    }
    // 5、将没计算完的情况进行计算，如只有1+2
    while(!s1.isEmpty()){
        cacl(s1 ,s2);
    }
    return s2.pop();
}

public void cacl(Stack<Character> s1 ,Stack<Integer> s2){
    if(s1.size()<1 || s2.size()<2){
        return;
    }
    char c=s1.pop();
    int n=s2.pop() ,m=s2.pop();
    if(c=='+'){
        s2.push(m+n);
    }else if(c=='-'){
        s2.push(m-n);
    }else{
        s2.push(m*n);
    }
}

class Solution {
  public:
    stack<int> num;
    stack<char> op;
    //优先级表
    map<char, int> h{ {'+', 1}, {'-', 1}, {'*', 2}};
    void eval() {
        int a = num.top();
        num.pop();
        int b = num.top();
        num.pop();
        int r = 0;
        if (op.top() == '+')
            r = a + b;
        if (op.top() == '-')
            r = b - a;
        if (op.top() == '*')
            r = a * b;
        op.pop();
        num.push(r);
    }
    int solve(string s) {
        for (int i = 0; i < s.size(); i++) {
            if (isdigit(
                        s[i])) { //数字入栈      isdigit(x)判断x是否是阿拉伯数字1~9
                int x = 0, j = i;//计算数字
                while (j < s.size() && isdigit(s[j])) {
                    x = x * 10 + s[j] - '0';
                    j++;
                }
                num.push(x);//数字入栈
                i = j - 1;
            }
            //左括号无优先级，直接入栈
            else if (s[i] == '(') { //左括号入栈
                op.push(s[i]);
            }
            //括号特殊，遇到左括号直接入栈，遇到右括号计算括号里面的
            else if (s[i] == ')') { //右括号
                while (op.top() != '(') //一直计算到左括号
                    eval();
                op.pop();//左括号出栈
            } else {
                while (op.size() &&
                        h[op.top()] >= h[s[i]])//待入栈运算符优先级低，则先计算
                    eval();
                op.push(s[i]);//操作符入栈
            }
        }
        while (op.size()) eval();//剩余的进行计算
        return num.top();
    }
};

非常好懂的，有注释喔

 function solve(s) {
  // write code here
  let ops = []; // 用来存储运算符
  let nums = []; // 用来存储数值和每次计算的结果
//   console.log(isNaN(parseInt(s[0])));
  for (let i = 0; i < s.length; i++) {
     if('(*'.indexOf(s[i]) > -1) { // 判断 s[i] 是否为  ( 和 * 
         ops.push(s[i])  // 是就入栈
     } else if(!isNaN(s[i])) { // 判断是否为 一个数字 ->number
         let temp = '' // 临时变量
         while(i<s.length && !isNaN(s[i])) {
             temp = temp + s[i++] // 因为 s 给的是一个字符串 所以通过这个办法 可以把 两位数的数字拼在一起
         }
         i-- // 如果 s[0] 和 s[1] 是一个操作数值12 经过上面的操作拼完了之后 i 会等于2 所以这里等让 i - 1 变回1 指向s[1]
         nums.push(parseInt(temp)) // 随后入栈
     } else if(s[i] == '+' || s[i] == '-') { // 如果是加号 或者 减号
         while(ops.length > 0 && '*+-'.indexOf(ops[ops.length - 1]) > -1) { // 就将 ops数组里
            //的  * + - 等运算符 pop 出去进行操作            
             let num1 = nums.pop()
             let num2 = nums.pop()
             let res = calc(ops.pop(), num1, num2)  // 加减乘 操作函数 在下面
             nums.push(res) // 将得出的结果入栈  ,  如果你有疑问， 为什么最后栈中的就一顶只有结果，没有别的操作数字， 因为
             // 上面 num1 和 num2 赋值的时候 都 pop出去了
         }
         ops.push(s[i]) // 最后将 此次遇到的 + 号丢进栈里
     } else if(s[i] == ')') { // 如果遇到 )
         while(ops.length > 0 && ops[ops.length - 1] != '(') { // 只要栈不空， 和不遇到 (
             // 就一直循环
             let num1 = nums.pop()
             let num2 = nums.pop()
             let res = calc(ops.pop(), num1, num2) // 思想和上面一样
             nums.push(res) // 结果入栈
         }
         ops.pop() // 把左括号丢出去
     }
  }
    while(ops.length > 0) {  // 最后 ops 不空 不停
        let num1=  nums.pop()
        let num2 = nums.pop()
        let temp_res = calc(ops.pop(), num1, num2)
        nums.push(temp_res) // 最后的结果 丢进去
    }
    return nums.pop() // 最后的结果 return 出去
}

function calc(op ,b ,a) {
    if(op == '+') return a + b
    if(op == '-') return a - b
    if(op == '*') return a * b
    return 0
}

module.exports = {
  solve: solve,
};

class Solution {
public:
    /**
     * 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
     */
    int solve(string s) {
        // write code here
        string cs;
        for (char &c : s) if (c != ' ') cs += c;
        int n = cs.size();
        stack<int> nums;
        nums.push(0);
        stack<char> ops;
        for (int i = 0; i < n; ++i) {
            char c = cs[i];
            if (c == '(') ops.push(c);
            else if (c == ')') {
                while (!ops.empty()) {
                    if (ops.top() != '(') calc(nums, ops);
                    else {
                        ops.pop();
                        break;
                    }
                }
            } else {
                if (isdigit(c)) {
                    int u = 0, j = i;
                    while (j < n && isdigit(cs[j])) u = u * 10 + (cs[j++] - '0');
                    nums.push(u);
                    i = j - 1;
                } else {
                    if (i > 0 && (cs[i - 1] == '(' || cs[i - 1] == '+' || cs[i - 1] == '-')) {
                        nums.push(0);
                    }
                    while (!ops.empty() && ops.top() != '(') {
                        char prev = ops.top();
                        if (m[prev] >= m[c]) {
                            calc(nums, ops);
                        } else {
                            break;
                        }
                    }
                    ops.push(c);
                }
            }
        }
        while (!ops.empty() && ops.top() != '(') calc(nums, ops);
        return nums.top();
    }
    void calc(stack<int> &nums, stack<char> &ops) {
        if (nums.empty() || nums.size() < 2) return;
        if (ops.empty()) return;
        int b = nums.top();
        nums.pop();
        int a = nums.top();
        nums.pop();
        char op = ops.top();
        ops.pop();
        int ans = 0;
        if (op == '+') ans = a + b;
        else if (op == '-') ans = a - b;
        else if (op == '*') ans = a * b;   
        nums.push(ans);
    }
    
    unordered_map<char, int> m = {{'-', 1}, {'+', 1}, {'*', 2}};
};

import java.util.*;
public class Solution {
    public int solve (String s) {
        // 单栈+递归
        char[] arr = s.replaceAll(" ","").toCharArray();
        LinkedList<Integer> stack=new LinkedList<>();
        int num=0;
        char sign='+';
        for(int i=0;i<arr.length;i++){
            char c=arr[i];
            if(Character.isDigit(c)){
                num=num*10+c-'0';
            }
            //遇到括号就截取括号内的字符串，再递归
            if(c=='('){
                int j=i+1;
                int flag=1;
                while(flag>0){
                    if(arr[j]=='(') flag++;
                    if(arr[j]==')') flag--;
                    j++;
                }
                num=solve(s.substring(i+1,j-1));
                i=j-1;
            }
            if(!Character.isDigit(c) || i==arr.length-1){
                //+-直接放，*/栈顶*num(遍历到的数字)再放
                if(sign == '+') stack.push(num);
                else if(sign == '-') stack.push(-1 * num);
                else if(sign == '*') stack.push(stack.pop() * num);
                else if(sign == '/') stack.push(stack.pop() / num);
                num=0;
                sign=c;
            }
        }
        int res=0;
        while(!stack.isEmpty()){
            res+=stack.pop();
        }
        return res;
    }
}

function solve( s ) {
    // write code here
    return ~~eval(s)
}

class Solution {
public:
    /**
     * 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
     */
    int solve(string s) {
        vector<string> nums;
        int len=s.size();
//         int pre=0;
        stack<char> sign;
        unordered_map<char,int> priority={{'(',0},{'+',1},{'-',1},{'*',2},{')',3}};
        for(int i=0;i<len;++i) {
            if(s[i]>='0' && s[i]<='9') {
                int j=i+1;
                while(j<len && s[j]>='0' && s[j]<='9')
                    ++j;
                nums.push_back(s.substr(i,j-i));
                i=j-1;
            } else {
                //运算符
                if(s[i]==')') {
                    while(sign.top()!='(') {
                        //将优先级小于等于当前符号的之前的符号出栈
                        nums.push_back(string(1,sign.top()));
                        sign.pop();
                    }
                    sign.pop();
                } else if(s[i]=='(') {
                    sign.push(s[i]);
                }else {
                    while(!sign.empty() && priority[sign.top()]>=priority[s[i]]) {
                        //将优先级大于等于当前符号的之前的符号出栈
                        nums.push_back(string(1,sign.top()));
                        sign.pop();
                    }
                    sign.push(s[i]);
                }
            }
        }
        while(!sign.empty()) {
            nums.push_back(string(1,sign.top()));
            sign.pop();
        }
        stack<int> res;
        for(string& cur: nums) {
            if(cur[0]>='0' && cur[0]<='9') {
                res.push(atoi(cur.c_str()));
            } else {
                int b=res.top(); res.pop();
                int a=res.top(); res.pop();
                switch(cur[0]) {
                    case '+': res.push(a+b); break;
                    case '-': res.push(a-b); break;
                    case '*': res.push(a*b); break;
                }
            }
        }
        return res.top();
    }
};

import java.util.*;


public class Solution {
    public int cal(int k2,int k1,char o){
        if(o=='+'){
            return k2+k1;
        }else if(o=='-'){
            return k2-k1;
        }else{
            return k2*k1;
        }
    }
    public int solve (String s) {
        //"(3+4)*(5+(2-3))"
        Stack<Integer> num = new Stack<>();
        Stack<Character> op = new Stack<>();
        int i = 0;
        while(i<s.length()){
            if(s.charAt(i)>='0' && s.charAt(i)<='9'){
                int k = 0;
                while(i<s.length() && s.charAt(i)>='0' && s.charAt(i)<='9'){
                    k = 10*k + (s.charAt(i)-'0');
                    i++;
                }
                num.push(k);
            }else{
                if(s.charAt(i)=='(')
                    op.push(s.charAt(i));
                else if(s.charAt(i)==')'){
                    while(op.peek()!='('){
                        int k1 = num.pop();
                        int k2 = num.pop();
                        char o = op.pop();
                        num.push(cal(k2,k1,o));
                    }
                    op.pop();
                }else if(s.charAt(i)=='*'){ //乘号不影响运算结果，不需要区分前一个运算符是加号还是乘号
                    op.push(s.charAt(i));
                }else{
                    while(num.size()>=2){
                        if(op.peek()=='(')//注意 运算符不能越过小括号
                            break;
                        int k1 = num.pop();
                        int k2 = num.pop();
                        char o = op.pop();
                        num.push(cal(k2,k1,o));
                    }
                    op.push(s.charAt(i));
                }
                i++;
            }
        }
        while(!op.isEmpty()){
            int k1 = num.pop();
            int k2 = num.pop();
            char o = op.pop();
            num.push(cal(k2,k1,o));
        }
        int res = num.pop();
        return res;
    }
}