本文主要是介绍LeetCode 题解（2）：Evaluate Reverse Polish Notation，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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
题解：
using namespace std;class Solution {
public:int evalRPN(vector<string> &tokens) {assert(!tokens.empty());stack<int> operand;for(vector<string>::iterator iter = tokens.begin(); iter < tokens.end(); iter++){const char* p = (*iter).c_str();if(p[1] == '\0' && ( p[0] == '+' || p[0] == '-' || p[0] == '*' || p[0] == '/' )){switch(*p){case 43:{assert(operand.size() > 1);int operand2 = operand.top();operand.pop();int operand1 = operand.top();operand.pop();operand.push(operand1 + operand2);break;}case 45:{assert(operand.size() > 1);int operand2 = operand.top();operand.pop();int operand1 = operand.top();operand.pop();operand.push(operand1 - operand2);break;}case 42:{assert(operand.size() > 1);int operand2 = operand.top();operand.pop();int operand1 = operand.top();operand.pop();operand.push(operand1 * operand2); break;}case 47:{assert(operand.size() > 1);int operand2 = operand.top();operand.pop();int operand1 = operand.top();operand.pop();operand.push(operand1 / operand2);break;}default:{break;}}}else{operand.push(atoi(p));}}assert(operand.size() == 1);return operand.top();}
};测试用例：
#include <vector>
#include <exception>
#include <stack>
#include <iostream>int main(int argc, char* argv [])
{string a[] = {"-78","-33","196","+","-19","-","115","+","-","-99","/","-18","8","*","-86","-","-","16","/","26","-14","-","-","47","-","101","-","163","*","143","-","0","-","171","+","120","*","-60","+","156","/","173","/","-24","11","+","21","/","*","44","*","180","70","-40","-","*","86","132","-84","+","*","-","38","/","/","21","28","/","+","83","/","-31","156","-","+","28","/","95","-","120","+","8","*","90","-","-94","*","-73","/","-62","/","93","*","196","-","-59","+","187","-","143","/","-79","-89","+","-"};vector<string> stringArray(a, a + sizeof(a)/sizeof(a[0]));Solution s;cout << s.evalRPN(stringArray) << endl;return 0;
}
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