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[AcWing 3302] 表达式求值

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#include<iostream>
#include<stack>
#include<cstring>
#include<unordered_map>

using namespace std;

stack<int> nums;
stack<char> op;
unordered_map<char, int> h{ {'+', 1}, {'-', 1}, {'*', 2}, {'/', 2} };
void eval()
{
    int b = nums.top();   nums.pop();
    int a = nums.top();   nums.pop();
    char p = op.top();  op.pop();
    int r = 0;
    if (p == '+')   r = a + b;
    if (p == '-')   r = a - b;
    if (p == '*')   r = a * b;
    if (p == '/')   r = a / b;
    nums.push(r);
}
int main()
{
    string s;
    cin >> s;
    for (int i = 0; i < s.size(); i ++) {
        if (isdigit(s[i])) {
            int x = 0, j = i;
            while (j < s.size() && isdigit(s[j])) {
                x = x * 10 + s[j] - '0';
                j ++;
            }
            nums.push(x);
            i = j - 1;
        }
        else if (s[i] == '(')   op.push(s[i]);
        else if (s[i] == ')') {
            while (op.top() != '(') eval();
            op.pop();
        }
        else {
            while (op.size() && h[op.top()] >= h[s[i]]) eval();
            op.push(s[i]);
        }
    }
    while (op.size())   eval();
    cout << nums.top();
    return 0;
}

标签:size,nums,int,top,pop,求值,AcWing,3302,op
来源: https://www.cnblogs.com/wKingYu/p/16209222.html