439. Tenary Expression Parser

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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:

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

Example 2:

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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:

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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"

解法1： Stack

打破传统思维，从后往前扫描放入stack
C++

1

Java

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public class Solution {
    public String parseTernary(String expression) {
        if (expression == null || expression.length() == 0) {
            return "";
        }
        Stack<Character> stack = new Stack<Character>();
        for (int i = expression.length() - 1; i >= 0; i--) {
            char current = expression.charAt(i);
            if (!stack.isEmpty() && stack.peek() == '?') {
                stack.pop();    // get ride of ?
                char left = stack.pop();
                stack.pop();
                char right = stack.pop();
                if (current == 'T') {
                    stack.push(left);
                } else {
                    stack.push(right);
                }
            } else {
                stack.push(current);
            }
        }
        return String.valueOf(stack.peek());
    }
}