Ternary Expression Parser Leetcode

Posted 璨璨要好好学习

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Ternary Expression Parser Leetcode相关的知识,希望对你有一定的参考价值。

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 0-9T 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"
 
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。。。= =

以上是关于Ternary Expression Parser Leetcode的主要内容,如果未能解决你的问题,请参考以下文章

Leetcode: Ternary Expression Parser

Ternary Expression Parser

Ternary Expression Parser Leetcode

PHP Cron Expression Parser ( LARAVEL )

Spring文档苦读Spring Expression Language(SpEL)

Ternary Tree