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  • LeetCode OJ

      题目:

      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

    解题思路:
      遇到数字则将其压入栈中,遇到运算符就弹出栈顶的两个元素做相应的运算,然后将运算结果压入栈中。
      最后的结果保存在栈顶。
    代码:
    class Solution {
    public:
        int evalRPN(vector<string> &tokens) {
            stack<int> st;
            int operand_a = 0, operand_b = 0, tmp;
            stringstream ss;
            vector<string>::iterator it;
            for(it = tokens.begin(); it != tokens.end(); it++){
                ss.clear();
                if ((*it) == "+" ){
                    operand_b = st.top(); st.pop();
                    operand_a = st.top(); st.pop();
                    st.push(operand_a + operand_b);
                }
                else if ((*it) == "-" ){
                    operand_b = st.top(); st.pop();
                    operand_a = st.top(); st.pop();
                    st.push(operand_a - operand_b);
                }
                else if ((*it) == "*" ){
                    operand_b = st.top(); st.pop();
                    operand_a = st.top(); st.pop();
                    st.push(operand_a * operand_b);
                }
                else if ((*it) == "/" ){
                    operand_b = st.top(); st.pop();
                    operand_a = st.top(); st.pop();
                    st.push(operand_a / operand_b);
                }
                else{
                    ss.str(*it);
                    ss >> tmp;
                    st.push(tmp);
                }
            }
            return st.top();
        }
    };


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  • 原文地址:https://www.cnblogs.com/dongguangqing/p/3726272.html
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