Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.
Note that after backspacing an empty text, the text will continue empty.
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 in python:
class Solution:
def backspaceCompare(self, S: str, T: str) -> bool:
stack1 = []
stack2 = []
for item in S:
if item != '#':
stack1.append(item)
else:
if stack1:
stack1.pop()
for item in T:
if item != '#':
stack2.append(item)
else:
if stack2:
stack2.pop()
print(stack1, stack2)
if len(stack1) != len(stack2):
return False
for i in range(len(stack1)):
if stack1[i] != stack2[i]:
return False
return True
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