439 Ternary Expression Parser

题目

ref: 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"

分析

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

代码

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




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