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:
The length of the given string is ≤ 10000.
Each number will contain only one digit.
The conditional expressions group right-to-left (as usual in most languages).
The condition will always be either T or F. That is, the condition will never be a digit.
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"

Solution：Stack

思路：
利用栈stack的思想，存所有的字符，将原数组从后往前遍历，将遍历到的字符都压入栈中，我们检测如果栈首元素是问号，说明我们当前遍历到的字符是T或F，然后我们移除问号，再取出第一部分，再移除冒号，再取出第二部分，我们根据当前字符来判断是放哪一部分进栈，这样遍历完成后，所有问号都处理完了，剩下的栈顶元素即为所求:
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

// Input: "T?T?F:5:3"
class Solution {
    public String parseTernary(String expression) {
        if (expression == null || expression.length() == 0) return "";
        Deque<Character> stack = new LinkedList<>();

        for (int i = expression.length() - 1; i >= 0; i--) {
            char c = expression.charAt(i);
            if (!stack.isEmpty() && stack.peek() == '?') {

                stack.pop(); //pop '?'
                char first = stack.pop();
                stack.pop(); //pop ':'
                char second = stack.pop();

                if (c == 'T') stack.push(first);
                else stack.push(second);
            } else {
                stack.push(c);
            }
        }

        return String.valueOf(stack.peek());
    }
}