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
orF
. That is, the condition will never be a digit. - The result of the expression will always evaluate to either a digit
0-9
,T
orF
.
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;
}
};