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Ternary Expression Parser Leetcode
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
TorF. That is, the condition will never be a digit. - The result of the expression will always evaluate to either a digit
0-9,TorF.
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"
这道题是用stack来解题的，是倒着来的，有的时候正着来不行可以倒着来嘛。倒着的主要原因是遇到问号就可以pop出需要判断的两个元素然后把结果放进去，正着不行是因为遇到问号的时候后面可能是一个表达式就有点尴尬了。。。
反正做题都要正反想一下！
public class Solution { public String parseTernary(String expression) { Stack<Character> s = new Stack<>(); char[] str = expression.toCharArray(); for (int i = str.length - 1; i >= 0; i--) { if (str[i] == ‘?‘) { char first = s.pop(); s.pop(); char second = s.pop(); if (str[i - 1] == ‘T‘) { s.push(first); } else { s.push(second); } i--; } else { s.push(str[i]); } } return String.valueOf(s.pop()); } }
到时候还是回顾下吧怕忘记了。。。= =
Ternary Expression Parser Leetcode
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