Lintcode: Expression Evaluation (Basic Calculator III)

Given an expression string array, return the final result of this expression

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Example
For the expression 2*6-(23+7)/(1+2),
input is

[
  "2", "*", "6", "-", "(",
  "23", "+", "7", ")", "/",
  (", "1", "+", "2", ")"
],
return 2

Note
The expression contains only integer, +, -, *, /, (, ).

这道题其实应该算Basic Calculator III. 参考了http://blog.csdn.net/nicaishibiantai/article/details/45740649

思路就是两个stack，一个存数字一个存符号。如果遇到数字直接存到数字stack；如果遇到符号，有几种情况：

1.当前符号比上一个符号优先级高，比如* 高于+，那么直接进栈

2.当前符号低于上一个，那么就要把所有已经在stack里面优先于当前符号的全算完，再推进当前符号

3.当前符号是“（”，直接push

4.当前符号是“）”，就要把所有“（”以前的符号全部算完

 public class Solution {
     /**
      * @param expression: an array of strings;
      * @return: an integer
      */
     public int evaluateExpression(String[] expression) {
         // write your code here
         Stack<Integer> integers = new Stack<Integer>();
         Stack<String> ops = new Stack<String>();
         int i = 0;
         while (i < expression.length) {
             String cur = expression[i];
             if (isOp(cur)) { // current string is an op
                 if (cur.equals("(")) ops.push(cur);
                 else if (cur.equals(")")) {
                     while (ops.size()>0 && !ops.peek().equals("(")) {
                         integers.push(calc(integers.pop(), integers.pop(), ops.pop()));
                     }
                     ops.pop();
                 }
                 else { // +,-,*,/
                     while (ops.size()>0 && precede(cur, ops.peek())) {
                         integers.push(calc(integers.pop(), integers.pop(), ops.pop()));
                     }
                     ops.push(cur);
                 }
             }
             else integers.push(Integer.parseInt(cur)); // current String is an integer, push to integer stack
             i++;
         }

         while (!ops.isEmpty()) {
             integers.push(calc(integers.pop(), integers.pop(), ops.pop()));
         }
         return integers.isEmpty()? 0 : integers.pop();
     }

     public boolean isOp(String input) {
         if (input.equals("+") || input.equals("-") || input.equals("*")
          || input.equals("/") || input.equals("(") || input.equals(")"))
             return true;
         return false;
     }

     public int calc(int a, int b, String op) {
         if (op.equals("+")) return a+b;
         else if (op.equals("-")) return b-a;
         else if (op.equals("*")) return a*b;
         else return b/a;
     }

     public boolean precede(String a, String b) {
         if (b.equals("*") || b.equals("/")) return true;
         if (b.equals("+") || b.equals("-")) {
             if (a.equals("*") || a.equals("/")) return false;
             else return true;
         }
         return false; //case like (a+b) 到第一个+号时，+和（比应该return false
     }
 };