LeetCode 227. 基本计算器 II

使用传统栈后缀表达式+弹栈计算，爆时间爆内存，垃圾解法；

mapmp;
vectorbs;

void init() {
	mp['+'] = 0;
	mp['-'] = 0;
	mp['*'] = 1;
	mp['/'] = 1;
}

void back_string(string s) {
	stacksign;
	int i = 0;
	while(i= '0' && s[i] <= '9') {
			string ss = "";
			while (s[i] >= '0' && s[i] <= '9') {
				ss += s[i];
				i++;
			}
			bs.push_back(ss);
		}
		else if(s[i]=='+'||s[i]=='-'||s[i]=='*'||s[i]=='/'){
			if (sign.empty()) {
				sign.push(s[i]);
			}
			else {
				if (mp[s[i]] > mp[sign.top()]) {
					sign.push(s[i]);
				}
				else {
					while (!sign.empty() && mp[s[i]] <= mp[sign.top()]) {
						string str = "";
						str += sign.top();
						bs.push_back(str);
						sign.pop();
					}
					sign.push(s[i]);
				}
			}
			i++;
		}
		else {
			i++;
		}
	}
	while (!sign.empty()) {
		string str = "";
		str += sign.top();
		bs.push_back(str);
		sign.pop();
	}
}

int calculate(string s) {
	init();
	stackst;
	back_string(s);
	int a, b;
	int ret = 0;
	for (int i = 0; i < bs.size(); i++) {
		if (bs[i] == "+") {
			a = st.top();
			st.pop();
			b = st.top();
			st.pop();
			st.push(a + b);
		}
		else if (bs[i] == "-") {
			a = st.top();
			st.pop();
			b = st.top();
			st.pop();
			st.push(b - a);
		}
		else if (bs[i] == "*") {
			a = st.top();
			st.pop();
			b = st.top();
			st.pop();
			st.push(a * b);
		}
		else if (bs[i] == "/") {
			a = st.top();
			st.pop();
			b = st.top();
			st.pop();
			st.push(b / a);
		}
		else {
			int cnt = 0;
			for (int j = 0; j < bs[i].size(); j++) {
				cnt = (bs[i][j] - '0') + cnt * 10;
			}
			st.push(cnt);
		}
	}
	return st.top();
}

最快的写法应该是采用之前简单加减运算之上，增加乘除优先级即可；

遇到加减元素，标记正负号之后按照数字入栈；

如果遇到乘除符号，则两位操作数只可能是现在的遍历得数目和栈顶元素，所以弹栈计算之后进行压栈；

但是值得注意的是，每次判断都是根据presign进行的，所以每次计算落后于实际遍历操作；

最后只需要对栈内元素进行加和计算即可；

class Solution {
public:
    int calculate(string s) {
        vector stk;
        char preSign = '+';
        int num = 0;
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            if (isdigit(s[i])) {
                num = num * 10 + int(s[i] - '0');
            }
            if (!isdigit(s[i]) && s[i] != ' ' || i == n - 1) {
                switch (preSign) {
                    case '+':
                        stk.push_back(num);
                        break;
                    case '-':
                        stk.push_back(-num);
                        break;
                    case '*':
                        stk.back() *= num;
                        break;
                    default:
                        stk.back() /= num;
                }
                preSign = s[i];
                num = 0;
            }
        }
        return accumulate(stk.begin(), stk.end(), 0);
    }
};