Evaluate Reverse Polish Notation

Evaluate Reverse Polish Notation

• 题目链接：http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/

　　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

• 题目大意：给定一个逆波兰表达式，求该表达式的值
• 思路：由于逆波兰表达式本身不需要括号来限制哪个运算该先进行，因此可以直接利用栈来模拟计算：遇到操作数直接压栈，碰到操作符直接取栈顶的2个操作数进行计算（注意第一次取出来的是右操作数），然后再把计算结果压栈，如此循环下去。最后栈中剩下的唯一一个元素便是整个表达式的值
• 实现：

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class Solution { public:     int evalRPN(vector<string> &tokens) {                   int result = 0;         int i;         stack<int> opd;         //存储操作数         int size = tokens.size();         for(i=0;i<size;i++)         {             if(tokens[i]=="*")             {                 int rOpd = opd.top();   //右操作数                 opd.pop();                 int lOpd = opd.top();  //左操作数                 opd.pop();                 result = lOpd*rOpd;                 opd.push(result);             }             else if(tokens[i]=="/")             {                 int rOpd = opd.top();                 opd.pop();                 int lOpd = opd.top();                 opd.pop();                 result = lOpd/rOpd;                 opd.push(result);             }             else if(tokens[i]=="+")             {                 int rOpd = opd.top();                 opd.pop();                 int lOpd = opd.top();                 opd.pop();                 result = lOpd+rOpd;                 opd.push(result);             }             else if(tokens[i]=="-")             {                 int rOpd = opd.top();                 opd.pop();                 int lOpd = opd.top();                 opd.pop();                 result = lOpd-rOpd;                 opd.push(result);             }             else             {                 opd.push(atoi(tokens[i].c_str()));             }         }         return opd.top();     } };