leetcode-Evaluate Reverse Polish Notation
2013-11-30 11:58
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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:
很基础的题。
int要用大整数方法。
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
很基础的题。
class Solution { public: int evalRPN(vector<string> &tokens) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. stack<int> st; int num1, num2; int i, j; for (i=0; i<tokens.size(); ++i) { if (tokens[i]=="+" || tokens[i]=="-" || tokens[i]=="*" || tokens[i]=="/") { num2 = st.top(); st.pop(); num1 = st.top(); st.pop(); st.push(cal(num1, num2, tokens[i])); } else { st.push(strToInt(tokens[i])); } } return st.top(); } private: int strToInt(const string &numStr) { int ret = 0; int i = 0, sign = 1; if (numStr[i] == '-') { sign = -1; ++i; } for (; i<numStr.size(); ++i) { ret = ret*10 + (numStr[i]-'0'); } if (sign==-1) ret = -ret; return ret; } int cal(int num1, int num2, const string &op) { int ret = 0; if (op == "+") { ret = num1 + num2; } else if (op == "-") { ret = num1 - num2; } else if (op == "*") { ret = num1 * num2; } else if (op == "/") { ret = num1 / num2; } return ret; } };但是还要考虑给出的逆波兰表达式是否合法,要在出栈那里判断是否有两个操作数以上。还有就是如果操作数大于
int要用大整数方法。
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