Evaluate Reverse Polish Notation
题意:
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
cpp:
class Solution { public: int evalRPN(vector<string>& tokens) { stack<int> stack; if (tokens.empty()) return 0; int len = tokens.size(); int a, b; for (int i = 0; i < len; i++) { string s = tokens[i]; if (s == "+") { a = stack.top(); stack.pop(); b = stack.top(); stack.pop(); stack.push(a + b); } else if (s == "-") { a = stack.top(); stack.pop(); b = stack.top(); stack.pop(); stack.push(b - a); } else if (s == "*") { a = stack.top(); stack.pop(); b = stack.top(); stack.pop(); stack.push(a * b); } else if (s == "/") { a = stack.top(); stack.pop(); b = stack.top(); stack.pop(); stack.push(b / a); } else { stack.push(atoi(s.c_str())); } } return stack.top(); } };
java:
public class Solution { public int evalRPN(String[] tokens) { String operators = "+-*/"; Stack<String> stack = new Stack<String>(); for (String s : tokens) { if (!operators.contains(s)) { stack.push(s); } else { int a = Integer.valueOf(stack.pop()); int b = Integer.valueOf(stack.pop()); int index = operators.indexOf(s); switch (index) { case 0: { stack.push(String.valueOf(a + b)); break; } case 1: { stack.push(String.valueOf(b - a)); break; } case 2: { stack.push(String.valueOf(a * b)); break; } case 3: { stack.push(String.valueOf(b / a)); break; } } } } return Integer.valueOf(stack.pop()); } };