Ternary Expression Parser

Posted 2020-08-26 北叶青藤

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"

 1 public class Solution {
 2     public String parseTernary(String expression) {
 3         if (expression == null || expression.length() == 0)
 4             return "";
 5         Stack<Character> stack = new Stack<>();
 6 
 7         for (int i = expression.length() - 1; i >= 0; i--) {
 8             char c = expression.charAt(i);
 9             if (!stack.isEmpty() && stack.peek() == ‘?‘) {
10 
11                 stack.pop(); // pop ‘?‘
12                 char first = stack.pop();
13                 stack.pop(); // pop ‘:‘
14                 char second = stack.pop();
15 
16                 if (c == ‘T‘)
17                     stack.push(first);
18                 else
19                     stack.push(second);
20             } else {
21                 stack.push(c);
22             }
23         }
24         return String.valueOf(stack.peek());
25     }
26 }