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
分析:這一題就是寫程序計算逆波蘭式的結果,遍歷表達式,碰到操作數入棧,碰到操作符就從棧頂取出兩個操作數,再將計算后的結果入棧,最后棧中剩余的唯一操作數就是計算結果。 本文地址
1 class Solution { 2 public: 3 int evalRPN(vector<string> &tokens) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 int len = tokens.size(); 7 stack<int> S; 8 for(int i = 0; i < len; i++) 9 { 10 if(tokens[i] == "+" || tokens[i] == "-" || 11 tokens[i] == "*" || tokens[i] == "/") 12 { 13 int op2 = S.top(); S.pop(); 14 int op1 = S.top(); S.pop(); 15 S.push( op(op1, op2, tokens[i][0]) ); 16 } 17 else 18 S.push(stoi(tokens[i])); 19 } 20 return S.top(); 21 } 22 int op(int op1, int op2, char optor) 23 { 24 if(optor == '+')return op1 + op2; 25 else if(optor == '-')return op1 - op2; 26 else if(optor == '*')return op1 * op2; 27 else return op1 / op2; 28 } 29 };
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