Leetcode: 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:

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"

 1 public class Solution {
 2     public String parseTernary(String expression) {
 3         Stack<String> st = new Stack<String>();
 4         int pos = expression.lastIndexOf("?");
 5         while (pos > 0) {
 6             if (pos < expression.length()-1) {
 7                 String str = expression.substring(pos+1);
 8                 String[] strs = str.split(":");
 9                 for (int i=strs.length-1; i>=0; i--) {
10                     if (strs[i].length() > 0)
11                         st.push(strs[i]);
12                 }
13             }
14             String pop1 = st.pop();
15             String pop2 = st.pop();
16             if (expression.charAt(pos-1) == ‘T‘) st.push(pop1);
17             else st.push(pop2);
18             expression = expression.substring(0, pos-1);
19             pos = expression.lastIndexOf("?");
20         }
21         return st.pop();
22     }
23 }

 1 public String parseTernary(String expression) {
 2     if (expression == null || expression.length() == 0) return "";
 3     Deque<Character> stack = new LinkedList<>();
 4 
 5     for (int i = expression.length() - 1; i >= 0; i--) {
 6         char c = expression.charAt(i);
 7         if (!stack.isEmpty() && stack.peek() == ‘?‘) {
 8 
 9             stack.pop(); //pop ‘?‘
10             char first = stack.pop();
11             stack.pop(); //pop ‘:‘
12             char second = stack.pop();
13 
14             if (c == ‘T‘) stack.push(first);
15             else stack.push(second);
16         } else {
17             stack.push(c);
18         }
19     }
20 
21     return String.valueOf(stack.peek());
22 }