LeetCode439. Ternary Expression

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.

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Example 1:

Input: "T?2:3"
Output: "2"
Explanation: If true, then result is 2; otherwise result is 3.

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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"

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Basic Idea:

Use Stack to process the whole String in a backward way. We process from back, to check the mark"?", if we meet "?", we take a look at the next condition, if true, we choose the result closed to the "?", if not, we choose the another one.

class Solution{
  public String parseTernary(String expression) {
      if (expression == null || expression.length() == 0) return "";//check corner cases.
      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();//the right answer
            stack.pop(); //pop ':'
            char second = stack.pop();//the false answer
            if (c == 'T') stack.push(first);
            else stack.push(second);
        } else {
            stack.push(c);
          }
    }
      return String.valueOf(stack.peek());
    }
}