439. Ternary Expression Parser

ref: https://leetcode.com/problems/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"

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要点应该就是从expression的从后往前处理，用一个stack记录每一个char，然后从后往前遇到某个字母，上一个字母是'?'的时候，就可以处理了，因为从后完全的第一个?应该第一个处理，然后如果是正确的表达式，走完的时候stack里面应该只剩下一个元素了，就是结果

    public String parseTernary(String expression) {
        if(expression == null || expression.length() == 0) {
            return "";
        }
        Deque<Character> stack = new ArrayDeque<>();
        for(int i = expression.length() - 1; i >= 0; i--) {
            char cur = expression.charAt(i);
            if(!stack.isEmpty() && stack.peekFirst() == '?') {
                stack.pollFirst();
                char trueOption = stack.pollFirst();
                stack.pollFirst();
                char falseOption = stack.pollFirst();
                if(cur == 'T') {
                    stack.offerFirst(trueOption);
                } else {
                    stack.offerFirst(falseOption);
                }
            } else {
                stack.offerFirst(cur);
            }
        }
        return stack.size() == 1? String.valueOf(stack.pollFirst()): "";
    }

时间复杂度是O(n), expression只走一遍