每日kata~05-Reverse-polish-notatio

逆波兰表达式

Your job is to create a calculator which evaluates expressions in Reverse Polish notation.

For example expression 5 1 2 + 4 * + 3 - (which is equivalent to 5 + ((1 + 2) * 4) - 3 in normal notation) should evaluate to 14.

Note that for simplicity you may assume that there are always spaces between numbers and operations, e.g. 1 3 + expression is valid, but 1 3+ isn't.

Empty expression should evaluate to 0.

Valid operations are +, -, *, /.

You may assume that there won't be exceptional situations (like stack underflow or division by zero).

Solution

正则&栈

第一次做的时候没判断栈为空的情况，加入判断之后√

做计算处理的时候有些麻烦了，用了最笨的方法(:з」∠)

public static double evaluate(String expr) {
        // TODO: Your awesome code here
      if(expr=="") {
          return 0;
      }
      String[] w = expr.split("\\s+");
      int l = w.length;
      double c = 0;

      Stack<String> st = new Stack<String>();

      for(int i=0;i<l;i++) {
          Pattern pattern = Pattern.compile("-?[0-9]+\\.?[0-9]*");
          Matcher isNum = pattern.matcher(w[i]);
          if(isNum.matches()) {
              st.push(w[i]);
          }
          else {
              double a = Double.parseDouble(st.pop());
              double b = Double.parseDouble(st.peek());
              if(!st.empty()) {
                  st.pop();
              }
              if((w[i].toCharArray())[0]=='+') {
                  c = b + a;
              }
              if((w[i].toCharArray())[0]=='-') {
                  c = b - a;
              }
              if((w[i].toCharArray())[0]=='*') {
                  c = b * a;
              }
              if((w[i].toCharArray())[0]=='/') {
                  c = b / a;
              }
              st.push(String.valueOf(c));
          }

      }            

        return Double.parseDouble(st.peek());
      }