leetcode_Evaluate Reverse Polish Notation

描述：

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

思路：

1.就像题目所描述的，计算逆波兰数学表达式的结果，循环访问字符串数组类型的表达式

2.遍历的字符是数字，将数字进栈，是数学符号的时，将连续的两个数字出栈并计算出结果，然后将结果再入栈

3.重复步骤2直至所有的数组元素被访问完毕。

代码：

//代码通俗易懂，略冗余。

public int evalRPN(String[] tokens) {        if(tokens==null||tokens.length==0)            return 0;        if(tokens.length==1)            return Integer.parseInt(tokens[0]);        int parseNum=0;        int num1=0,num2=0;        Stack<Integer>st=new Stack<Integer>();        for(int i=0;i<tokens.length;i++)//iterate the item of the array        {        if(isOperator(tokens[i]))//if the item is operator,caculate the numbers        {        num2=st.peek();        st.pop();        num1=st.peek();        st.pop();        parseNum=evaluteNums(num1,num2,tokens[i]);        if(i+1==tokens.length)        return parseNum;        st.push(parseNum);        }else {st.push(Integer.parseInt(tokens[i]));//if the item is number,push to the stack}                }        return parseNum;    }public int evaluteNums(int num1,int num2,String operator){if(operator.equals("+"))return num1+num2;else if(operator.equals("-"))return num1-num2;else if(operator.equals("*"))return num1*num2;else return num1/num2;}public boolean isOperator(String str){char operator=str.charAt(0);if(operator=='+'||operator=='-'||operator=='*'||operator=='/'){if(str.length()==1)return true;}return false;}