LeetCode227. Basic Calculator II

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

• You may assume that the given expression is always valid.
• Do not use the eval built-in library function.

思路：

解法1

我们是如同第224题，通过使用两个stack来分别存放数字和符号，但是因为乘除的优先级高于加减，所有我们先处理乘除，然后再处理加减。

class Solution {
public:
    int calculate(string s) {
		list<int> nums, signs;
		for (int i = 0; i<int(s.size()); i++) {
			if (isdigit(s[i])) {
				nums.push_back(getNext(s, i));
			}
			else if (s[i] == '-') signs.push_back(-1);
			else if (s[i] == '+') signs.push_back(1);
			else if (s[i] == '*') {
				while (!isdigit(s[i])) i++;
				nums.back() *= getNext(s, i);
			}
			else if (s[i] == '/') {
				while (!isdigit(s[i])) i++;
				nums.back() /= getNext(s, i);
			}
		}

		int sum = 0;
		signs.push_front(1);
		list<int>::iterator itNum = nums.begin(), itSign = signs.begin();
		while (itNum != nums.end()) {
			sum += (*itNum) * (*itSign);
            itNum++;itSign++;
		}
		return sum;
    }
private:
	int getNext(const string& s, int& i) {
		string numS;
		while (isdigit(s[i])) {
			numS += s[i++];
		}
		i--;
		return stoi(numS);
	}
};

解法2：

我们每次都将操作符之前的一个数和操作符存下来，这样当我们遇到乘除运算的时候可以优先处理，也可以一次处理完所有的数据。

class Solution {
public:
    int calculate(string s) {
		long pre = 0, sum = 0, curr = 0;
		char opt = '+';
		int n = int(s.size());
		for (int i = 0; i < n; i++) {
			if (isdigit(s[i])) {
				curr = curr * 10 + s[i] - '0';
			}
			if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/' || i == n-1) {
				switch (opt) {
				case '+': pre += curr; break;
				case '-': pre -= curr; break;
				case '*': pre *= curr; break;
				case '/': pre /= curr; break;
				default: break;
				}
				if (s[i] == '-' || s[i] == '+' || i == n - 1) {
					sum += pre;
					pre = 0;
				}
				curr = 0;
				opt = s[i];
			}
		}
		return int(sum);
    }
};