LEETCODE: Evaluate Reverse Polish Notation

最新推荐文章于 2019-07-28 09:56:10 发布

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最新推荐文章于 2019-07-28 09:56:10 发布

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文章标签： LEETCODE 算法 stack

本文链接：https://blog.youkuaiyun.com/anyicheng2015/article/details/42675931

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Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Some examples:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

class Solution {
public:
    
    bool isFlag(string str) {
        if(str.length() != 1)
            return false;
        if(str[0] == '-' || str[0] == '+' || str[0] == '/' || str[0] == '*')
            return true;
        return false;
    }
    
    int calculate(int left, int right, string flag) {
        if(flag[0] == '+')
            return left + right;
        if(flag[0] == '-')
            return left - right;
        if(flag[0] == '*')
            return left * right;
        if(flag[0] == '/')
            return left / right;
        return 0;
    }
    
    int stringToNumber(string str) {
        int number = 0;
        int flag = 1;
        for(int ii = 0; ii < str.length(); ii ++) {
            if(ii == 0) {
                if(str[0] == '-') {
                    flag = -1;
                    continue;
                }
                else if(str[0] == '+') {
                    continue;
                }
            }
            number = number * 10 + str[ii] - '0';
        }
        return number * flag;
    }

    int evalRPN(vector<string> &tokens) {
        if(tokens.size() == 0)
            return 0;

        stack<int> st;
        for(int ii = 0; ii < tokens.size(); ii ++) {
            if(isFlag(tokens[ii])) {
                int right = st.top();
                st.pop();
                int left = st.top();
                st.pop();
                int current  = calculate(left, right, tokens[ii]);
                st.push(current);
            }
            else {
                st.push(stringToNumber(tokens[ii]));
            }
        }
        return st.top();
    }
};