439. Ternary Expression Parser

Given a string representing arbitrarily nested ternary expressions, calculate the result of the expression. You can always assume that the given expression is valid and only consists of digits 0-9, ?, :, T and F (T and F represent True and False respectively).

Note:

1. The length of the given string is ≤ 10000.
2. Each number will contain only one digit.
3. The conditional expressions group right-to-left (as usual in most languages).
4. The condition will always be either T or F. That is, the condition will never be a digit.
5. The result of the expression will always evaluate to either a digit 0-9, T or F.

Example 1:

Input: "T?2:3"

Output: "2"

Explanation: If true, then result is 2; otherwise result is 3.

Example 2:

Input: "F?1:T?4:5"

Output: "4"

Explanation: The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:

             "(F ? 1 : (T ? 4 : 5))"                   "(F ? 1 : (T ? 4 : 5))"
          -> "(F ? 1 : 4)"                 or       -> "(T ? 4 : 5)"
          -> "4"                                    -> "4"

Example 3:

Input: "T?T?F:5:3"

Output: "F"

Explanation: The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:

             "(T ? (T ? F : 5) : 3)"                   "(T ? (T ? F : 5) : 3)"
          -> "(T ? F : 3)"                 or       -> "(T ? F : 5)"
          -> "F"                                    -> "F"

从右往左看，这个表达式我们可以试着看一看。
每次计算都是以‘?’ 为标志，当我们碰到'?'的时候，我们计算local result，并不需要区分trueStack VS. falseStack 因为我们碰到?的时候只需要看最近的三个点就可以了。即使是
?():() 这样的表达式，因为我们已经算过local的结果，因此还是比较最近的三个点。

 public String parseTernary(String expression) {  
     // why only judge on ?  
     // we look from right to left, think how we do it when we compute  
     // we ignore things except '?', when we meet '?', we look at nearby three elements  
     // event through ?():(), this situation, we've already calculated result for things inside()  
     // use stack to preserve previous solutions  
     Stack<Character> stack = new Stack<>();  
     for(int i = expression.length()-1; i>=0; i--){  
       if(expression.charAt(i) == '?'){  
         char t = stack.pop();  
         stack.pop();  
         char f = stack.pop();  
         if(expression.charAt(i-1) == 'F'){  
           stack.push(f);  
         }else {  
           stack.push(t);  
         }  
         i--;  
       }else{  
         stack.push(expression.charAt(i));  
       }  
     }  
     return stack.pop()+"";  
   }