[Leetcode] 439. Ternary Expression Parser 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/78479290

本文介绍了一种解析复杂三元表达式的算法实现，通过递归和迭代两种方式详细阐述了如何正确计算这类表达式的值，并附带示例说明。

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题目：

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"

思路：

由于右边的Ternary expression需要优先被parse，所以我们每次先找到最右边的‘？’进行parse，直到整个Ternary expression被完全parse为止。值得注意的是，虽然递归版本的思路更为直观，但是在本题目的测试中会造成内存耗尽，我改成迭代版本才通过测试。

代码：

1、递归版本：

class Solution {
public:
    string parseTernary(string expression) {
        if (expression.length() == 5) {
            char c = expression[0] == 'T' ? expression[2] : expression[4];
            string ret;
            ret += c;
            return ret;
        }
        else {                         // find the last '?' and then parse it
            int index = expression.find_last_of('?');
            string new_expression = expression.substr(0, index - 1);
            new_expression += parseTernary(expression.substr(index - 1, 5));
            if (index + 5 < expression.length()) {
                new_expression += expression.substr(index + 4);
            }
            return parseTernary(new_expression);
        }
    }
};

2、迭代版本：

class Solution {
public:
    string parseTernary(string expression) {
        while (expression.length() > 1) {
            int index = expression.find_last_of('?');   // find the last '?' and then parse it
            string new_expression;
            if (index - 1 > 0) {
                new_expression += expression.substr(0, index - 1);
            }
            new_expression += (expression[index - 1] == 'T' ? expression[index + 1] : expression[index + 3]);
            if (index + 5 < expression.length()) {
                new_expression += expression.substr(index + 4);
            }
            expression = new_expression;
        }
        return expression;
    }
};