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EvaluateReversePolishNotation

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	`1`	`+/*`
	`2`	`+ Author: Andy, nkuwjg@gmail.com`
	`3`	`+ Date: Jan 31, 2015`
	`4`	`+ Problem: Evaluate Reverse Polish Notation`
	`5`	`+ Difficulty: Easy`
	`6`	`+ Source: https://oj.leetcode.com/problems/evaluate-reverse-polish-notation/`
	`7`	`+ Notes:`
	`8`	`+ Evaluate the value of an arithmetic expression in Reverse Polish Notation.`
	`9`	`+ Valid operators are +, -, *, /. Each operand may be an integer or another expression.`
	`10`	`+ Some examples:`
	`11`	`+ ["2", "1", "+", "3", ""] -> ((2 + 1) 3) -> 9`
	`12`	`+ ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6`
	`13`	`+ Solution: stack.`
	`14`	`+ */`
	`15`	`+`
	`16`	`+publicclassSolution {`
	`17`	`+publicintevalRPN(String[]tokens) {`
	`18`	`+Stack<Integer>stk =newStack<Integer>();`
	`19`	`+for (inti =0;i <tokens.length; ++i) {`
	`20`	`+if ((tokens[i].compareTo("+") !=0) && (tokens[i].compareTo("-") !=0)`
	`21`	`+ && (tokens[i].compareTo("*") !=0) && (tokens[i].compareTo("/") !=0)) {`
	`22`	`+stk.push(Integer.parseInt(tokens[i]));`
	`23`	`+ }else {`
	`24`	`+intop2 =stk.pop();`
	`25`	`+intop1 =stk.pop();`
	`26`	`+intres =0;`
	`27`	`+if(tokens[i].compareTo("+") ==0)`
	`28`	`+res =op1 +op2;`
	`29`	`+elseif(tokens[i].compareTo("-") ==0)`
	`30`	`+res =op1 -op2;`
	`31`	`+elseif(tokens[i].compareTo("*") ==0)`
	`32`	`+res =op1 *op2;`
	`33`	`+elseif(tokens[i].compareTo("/") ==0)`
	`34`	`+res =op1 /op2;`
	`35`	`+stk.push(res);`
	`36`	`+ }`
	`37`	`+ }`
	`38`	`+returnstk.pop();`
	`39`	`+ }`
	`40`	`+}`

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