题目

• 请写一个整数计算器，支持加减乘三种运算和括号。
  • 示例1
    输入：“1+2”
    返回值：3
  • 示例2
    输入：“(2*(3-4))*5”
    返回值：-10
  • 示例3
    输入：“3+2*3*4-1”
    返回值：26

思路

• 经典的中缀表达式求值。常用思路，将其转为后缀表达式（后缀表达式的特点：遇到运算符则将前两个数拿出来运算，便于计算机计算），边转边求值。
• 读到左括号 ( 较为特殊，读到左括号时直接入栈，
• 本题中运算符优先级：左括号或栈顶为空、乘、加减 、右括号。
• 读到右括号时一直出栈直到一个左括号出栈。

算法步骤

• 遍历字符串，当 遍历到的字符是数字 时，直接将其压入数字栈。
• 否则，当 遍历到的字符是运算符 时，判断字符，直到字符串读完：
  • 若符号栈为空，则直接将其压入符号栈。
  • 若符号栈非空，则判断符号栈栈顶的运算符优先级：
    • 若 遍历到的运算符符号的优先级 > 符号栈栈顶元素的优先级，或符号栈空，则将读取的运算符压入符号栈。
    • 若 遍历到的运算符符号的优先级 ≤ 符号栈栈顶元素的优先级，则将其出栈，并从数字栈弹出两个元素进行该运算，先弹出的为右操作数，后弹出的为左操作数，将结果压入数字栈。
• 最后，若符号栈非空，则弹出栈顶元素，并从数字栈弹出两个元素来进行该运算，重复该操作直到符号栈空。
• 符号栈为空后，弹出数字栈栈顶元素作为结果。
• 返回数值站栈顶元素作为最终结果。

int solve(string s) {stack<char> sign;stack<int> num;for (int i = 0; i < s.length(); i++) {//若当前遍历到的字符为数字，则将连续的数放入val栈if (s[i] >= '0' && s[i] <= '9') {int rear = i;while ((rear + 1) < s.length() && s[rear + 1] >= '0' && s[rear + 1] <= '9') {rear++;}int n = 0;for (int j = rear; j >= i; j--) {n += (s[j] - '0') * pow(10, (rear - j));}i = rear;num.push(n);}//【优先级较大】当前遍历到的字符为(、*  或  ops符号栈为空，则将该符号压入符号栈else if (s[i] == '(' || s[i] == '*'|| sign.size() == 0   ){sign.push(s[i]);}//【优先级较小】当前遍历到的字符为+、-else if (s[i] == '+' || s[i] == '-') {while (1) {if (sign.size() == 0)break;if (sign.top() == '*') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 * n2);}else if (sign.top() == '+') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 + n2);}else if (sign.top() == '-') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 - n2);}else {break;}}sign.push(s[i]);}// 读到右括号时符号栈就一直出栈直到一个左括号出栈。else if (s[i] == ')') {while (1) {if (sign.size() == 0) {break;}else if (sign.top() == '(') {sign.pop();break;}else if (sign.top() == '+') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 + n2);}else if (sign.top() == '-') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 - n2);}else if (sign.top() == '*') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 * n2);}}}}//最后，若符号栈非空，则弹出栈顶元素，并从数字栈弹出两个元素来进行该运算。重复该操作直到符号栈空。while (sign.size() != 0) {if (sign.top() == '+') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 + n2);}else if (sign.top() == '-') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 - n2);}else if (sign.top() == '*') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 * n2);}}//弹出数字栈栈顶元素，即得结果。return num.top();
}

实例

#include <iostream>
#include <stack>
using namespace std;int solve(string s) {// write code herestack<char> sign;stack<int> num;for (int i = 0; i < s.length(); i++) {//若当前遍历到的字符为数字，则将连续的数放入val栈if (s[i] >= '0' && s[i] <= '9') {int rear = i;while ((rear + 1) < s.length() && s[rear + 1] >= '0' && s[rear + 1] <= '9') {rear++;}int n = 0;for (int j = rear; j >= i; j--) {n += (s[j] - '0') * pow(10, (rear - j));}i = rear;num.push(n);}//若 当前遍历到的字符为(、*  或  ops符号栈为空，则将该符号压入符号栈else if (s[i] == '(' || s[i] == '*'|| sign.size() == 0   ){sign.push(s[i]);}//若 当前遍历到的字符为+、-else if (s[i] == '+' || s[i] == '-') {while (1) {if (sign.size() == 0)break;if (sign.top() == '*') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 * n2);}else if (sign.top() == '+') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 + n2);}else if (sign.top() == '-') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 - n2);}else {break;}}sign.push(s[i]);}else if (s[i] == ')') {while (1) {if (sign.size() == 0) {break;}else if (sign.top() == '(') {sign.pop();break;}else if (sign.top() == '+') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 + n2);}else if (sign.top() == '-') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 - n2);}else if (sign.top() == '*') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 * n2);}}}}//最后，若符号栈非空，则弹出栈顶元素，并从数字栈弹出两个元素来进行该运算。重复该操作直到符号栈空。while (sign.size() != 0) {if (sign.top() == '+') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 + n2);}else if (sign.top() == '-') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 - n2);}else if (sign.top() == '*') {sign.pop();int n2 = num.top();num.pop();int n1 = num.top();num.pop();num.push(n1 * n2);}}//弹出数字栈栈顶元素，即得结果。return num.top();
}int main()
{string mys = "1+3*(5+2)";cout<< mys<<"=" << solve(mys) << endl;return 0;
}