439 Ternary Expression Parser

439. Ternary Expression Parser​
1. Question
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 digits0-9,?,:,TandF(TandFrepresent 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 eitherTorF. That is, the condition will never be a digit.
5.The result of the expression will always evaluate to either a digit0-9,TorF.
Example 1:
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Input:"T?2:3"
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Output:"2"
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Explanation:If true, then result is 2; otherwise result is 3.
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Example 2:
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Input:"F?1:T?4:5"
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Output:"4"
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​
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Explanation: The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:
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​
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             "(F ? 1 : (T ? 4 : 5))"                   "(F ? 1 : (T ? 4 : 5))"
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           ->"(F ? 1 : 4)"                 or       -> "(T ? 4 : 5)"
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           ->"4"                                    -> "4"
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Example 3:
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Input:"T?T?F:5:3"
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Output:"F"
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Explanation: The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:
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​
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             "(T ? (T ? F : 5) : 3)"                   "(T ? (T ? F : 5) : 3)"
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          -> "(T ? F : 3)"                 or       -> "(T ? F : 5)"
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          -> "F"                                    -> "F"
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2. Implementation
(1) Stack
思路：这是当年面Pocket Gem的电面题，当时的我完全懵逼无从入手...这道题既可以用递归，也可以用stack解。如果用stack的话，需要从后往前扫，这样的好处是当我们遇到'?'前面的character(T 或者 F)，我们可以知道该保留‘?’后面的哪个expression
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class Solution {
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    public String parseTernary(String expression) {
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        if (expression == null || expression.length() == 0) {
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            return "";
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        }
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​
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        Stack<Character> stack = new Stack<>();
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​
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        for (int i = expression.length() - 1; i >= 0; i--) {
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            char c = expression.charAt(i);
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​
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            if (!stack.isEmpty() && stack.peek() == '?') {
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                stack.pop();
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​
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                char firstVal = stack.pop();
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                stack.pop();
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                char secondVal = stack.pop();
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​
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                if (c == 'T') {
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                    stack.push(firstVal);
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                }
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                else {
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                    stack.push(secondVal);
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                }
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            }
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            else {
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                stack.push(c);
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            }
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        }
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        return String.valueOf(stack.pop());
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    }
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}
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(2) DFS
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class Solution {
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    public String parseTernary(String expression) {
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        int len = expression.length();
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​
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        if (len == 1) {
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            return expression;
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        }
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​
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        int count = 0;
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        for (int i = 0; i < len; i++) {
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            if (expression.charAt(i) == '?') {
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                ++count;
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            }
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            else if (expression.charAt(i) == ':') {
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                --count;
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​
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                // When the number of '?' matches with ':', we get a valid expression
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                if (count == 0) {
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                    if (expression.charAt(0) == 'T') {
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                        // Get the expression before ':', skip 'T' and '?' at the beginning
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                        return parseTernary(expression.substring(2, i));
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                    } 
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                    else {
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                        // Get the expression after ':'
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                        return parseTernary(expression.substring(i + 1));
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                    }
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                }
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            }
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        }
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        return "";
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    }
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}
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3. Time & Space Complexity
Stack: 时间复杂度O(n), 空间复杂度O(n)
DFS: 时间复杂度O(n), 空间复杂度O(n)

402 Remove K Digits

456 132 Pattern

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Contents

439. Ternary Expression Parser

1. Question

2. Implementation

3. Time & Space Complexity

439. Ternary Expression Parser​

1. Question

2. Implementation

3. Time & Space Complexity

439. Ternary Expression Parser