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
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题目是用 stack 维护一个公式的进出,判断方法还可以,开始忘记数可能为负,后面改进了。
#include <stack> #include <iostream> #include <string> #include <vector> using namespace std; class Solution { public: int evalRPN(vector<string> &tokens) { int n = tokens.size(); stack<int > tmp; for(int i=0;i<n;i++){ if(tokens[i][0]>=‘0‘&&tokens[i][0]<=‘9‘){ tmp.push( helpFun(tokens[i]) ); continue; } if(tokens[i][0]==‘-‘&&tokens[i][1]!=‘\0‘){ tmp.push( helpFun( tokens[i])); continue; } int rgt = tmp.top(); tmp.pop(); int lft = tmp.top(); tmp.pop(); switch (tokens[i][0]){ case ‘+‘: tmp.push( lft + rgt ); break; case ‘-‘: tmp.push(lft - rgt); break; case ‘*‘: tmp.push(lft * rgt); break; case ‘/‘: tmp.push(lft / rgt); break; } } return tmp.top(); } int helpFun(string str) { int sum = 0,i = 0; if (str[0]==‘-‘) i = 1; for(;i<str.length();i++) sum = sum*10+str[i]-‘0‘; return str[0]==‘-‘?-sum:sum; } }; int main() { vector<string> tokens{"3","-4","+"}; Solution sol; cout<<sol.evalRPN(tokens)<<endl; // for(int i=0;i<tokens.size();i++) // cout<<tokens[i]<<endl; return 0; }
时间: 2024-07-31 14:21:15