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"

queue存储，从后向前遍历字符串，遇到peek = ？，pop出来？，pop出来first，pop出来：，pop出来second。如果当前是T，push first，否则push second。代码如下：

public class Solution {    public String parseTernary(String expression) {        if (expression == null || expression.length() == 0) {            return "";        }        Deque<Character> stack = new LinkedList<Character>();        for (int i = expression.length() - 1; i >= 0; i --) {            char ch = expression.charAt(i);            if (!stack.isEmpty() && stack.peek() == '?') {                stack.pop(); //'?'                char first = stack.pop();                stack.pop(); //':'                char second = stack.pop();                stack.push(ch == 'T'? first: second);            } else {                stack.push(ch);            }        }        return stack.pop() + "";    }}