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[LeetCode]-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
1 /* 2 *解题思路： 3 *借助辅助栈来完成逆波兰表达式的计算 4 *1.遍历数组，当前字符串为数字时入栈； 5 *2.为算符时，弹出栈顶的两个元素，完成计算，计算结果入栈； 6 *3.遍历完成，当栈非空时，返回栈顶元素，计算完成。 7 */ 8 public class Solution { 9 public int evalRPN(String[] tokens) {10 int factor1, factor2, result=0;11 Stack<String> stack = new Stack<String>();12 int len = tokens.length;13 for( int i=0; i<len; i++ ){14 if( tokens[i].equals("+") ){15 factor1 = Integer.parseInt( stack.pop() );16 factor2 = Integer.parseInt( stack.pop() );17 result = factor1 + factor2;18 stack.push(result+"");19 20 }else if( tokens[i].equals("-") ){21 factor1 = Integer.parseInt( stack.pop() );22 factor2 = Integer.parseInt( stack.pop() );23 result = factor2 - factor1;24 stack.push(result+"");25 26 }else if( tokens[i].equals("*") ){27 factor1 = Integer.parseInt( stack.pop() );28 factor2 = Integer.parseInt( stack.pop() );29 result = factor1 * factor2;30 stack.push(result+"");31 32 }else if( tokens[i].equals("/") ){33 factor1 = Integer.parseInt( stack.pop() );34 factor2 = Integer.parseInt( stack.pop() );35 result = factor2 / factor1;36 stack.push(result+"");37 38 }else{39 stack.push( tokens[i] );40 }41 }42 43 if( !stack.isEmpty() )44 result = Integer.parseInt( stack.pop() );45 return result; 46 }47 }
[LeetCode]-Evaluate Reverse Polish Notation
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