leetcode 224. 基本計算器(逆波蘭表示式)

題目

不能處理負數，蕪湖。比如3*(-5).

class Solution {
#define ll long long
public:
    int getprio(int x){
        if(x=='('-'0') return 1;//左括號在這個演算法裡優先順序最低
        if(x=='*'-'0'||x=='/'-'0') return 3;
        if(x=='+'-'0'||x=='-'-'0') return 2;
        return 0;
    }
    int calculate(string s) {
        int n=s.size();
        if(!n) return 0;
        //得到字尾表示式,並儲存在s2中
        stack<int>s1,s2;
        for(int i=0,j;i<n;i=j){//s[i]不是數字
            if(s[i]==' ') {j=i+1;continue;}//空格
            if(!(s[i]>='0'&&s[i]<='9')){
                j=i+1;
                if(s[i]=='(') s1.push(s[i]-'0');//s[i]是左括號
                else if(s[i]==')'){//s[i]是右括號
                    while(s1.top()!='('-'0') s2.push(s1.top()),s1.pop();
                    s1.pop();
                }else{//+-*/
                    while(!s1.empty()&&getprio(s1.top())>=getprio(s[i]-'0')) s2.push(s1.top()),s1.pop();
                    s1.push(s[i]-'0');
                }
            }else{//s[i]是數字
                j=i;
                int tmp=0;
                while(j<n&&(s[j]>='0'&&s[j]<='9'||s[j]==' ')) (s[j]==' ')?(++j):(tmp=tmp*10+(ll)s[j++]-'0');//這裡由於是char型別,s[j]可能會爆int
                s2.push(tmp);
            }
        }
        while(!s1.empty()) s2.push(s1.top()),s1.pop();
        while(!s2.empty()) s1.push(s2.top()),s2.pop();
        //計算字尾表示式s1,s2作為計算的輔助棧
        while(!s1.empty()){
            if(s1.top()>=0) s2.push(s1.top());
            else{
                int y=s2.top();s2.pop();
                int x=s2.top();s2.pop();
                if(s1.top()=='+'-'0') s2.push(x+y);
                if(s1.top()=='-'-'0') s2.push(x-y);
                if(s1.top()=='*'-'0') s2.push(x*y);
                if(s1.top()=='/'-'0') s2.push(x/y);
            }
            s1.pop();
        }
        return s2.top();
    }
};