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
,?
,:
,T
andF
(T
andF
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"
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
class Solution {
public String parseTernary(String expression) {
if (expression == null || expression.length() == 0) {
return "";
}
Stack<Character> stack = new Stack<>();
for (int i = expression.length() - 1; i >= 0; i--) {
char c = expression.charAt(i);
if (!stack.isEmpty() && stack.peek() == '?') {
stack.pop();
char firstVal = stack.pop();
stack.pop();
char secondVal = stack.pop();
if (c == 'T') {
stack.push(firstVal);
}
else {
stack.push(secondVal);
}
}
else {
stack.push(c);
}
}
return String.valueOf(stack.pop());
}
}
(2) DFS
class Solution {
public String parseTernary(String expression) {
int len = expression.length();
if (len == 1) {
return expression;
}
int count = 0;
for (int i = 0; i < len; i++) {
if (expression.charAt(i) == '?') {
++count;
}
else if (expression.charAt(i) == ':') {
--count;
// When the number of '?' matches with ':', we get a valid expression
if (count == 0) {
if (expression.charAt(0) == 'T') {
// Get the expression before ':', skip 'T' and '?' at the beginning
return parseTernary(expression.substring(2, i));
}
else {
// Get the expression after ':'
return parseTernary(expression.substring(i + 1));
}
}
}
}
return "";
}
}
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)
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