标签：problem amp wiki 波兰表示法 ble truncate ESS use string
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Note:
Example 1:
Input: ["2", "1", "+", "3", "*"] Output: 9 Explanation: ((2 + 1) * 3) = 9
思路
I would use stack to help solving this problem
Traverse the whole given string array
1. if operand is integer, push into stack
2. if operand is operation, pop two items from stack, do caculation and push result back to stack
代码
1 class Solution { 2 public int evalRPN(String[] tokens) { 3 Stack<Integer> stack = new Stack<>(); 4 for (String s : tokens) { 5 if(s.equals("+")) { 6 stack.push(stack.pop()+stack.pop()); 7 }else if(s.equals("/")) { 8 int latter = stack.pop(); 9 int former = stack.pop(); 10 stack.push( former / latter); 11 }else if(s.equals("*")) { 12 stack.push(stack.pop() * stack.pop()); 13 }else if(s.equals("-")) { 14 int latter = stack.pop(); 15 int former = stack.pop(); 16 stack.push(former - latter); 17 }else { 18 stack.push(Integer.parseInt(s)); 19 } 20 } 21 return stack.pop(); 22 } 23 }
[leetcode]150. Evaluate Reverse Polish Notation逆波兰表示法
标签：problem amp wiki 波兰表示法 ble truncate ESS use string
原文地址：https://www.cnblogs.com/liuliu5151/p/9901121.html
