Given two strings S and T, return if they are equal when both are typed into empty text editors. #
means a backspace character.
Example 1:
Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".
Example 2:
Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".
Example 3:
Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".
Example 4:
Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".
Note:
1 <= S.length <= 200
1 <= T.length <= 200
- S and T only contain lowercase letters and
'#'
characters.
Follow up:
Can you solve it in O(N)
time and O(1)
space?
Solution
Just use two stacks to resolve, then compare result
class Solution {
func backspaceCompare(_ S: String, _ T: String) -> Bool {
return resolve(S) == resolve(T)
}
func resolve(_ str: String) -> String {
var result = ""
for c in str {
if c != "#" { result.append(c) }
else {
if result.isEmpty { continue }
result.removeLast()
}
}
return result
}
}
For
O(1)
space, I'm thinking about move backward...