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 <= 2001 <= T.length <= 200-
SandTonly contain lowercase letters and'#'characters.
Follow up:
- Can you solve it in
O(N)time andO(1)space?
建一个helper function, 然后用stack, 如果为'#', 就stack.pop(), 否则append(c), 最后返回"".join(stack). T: O(m+n) S: O(m+n)
**imporve, 可以用two pointers 去实现 T: O(m+n) S: O(1)
Code
class Solution:
def backspaceStringCompare(self, S, T):
def helper(s):
stack = []
for c in s:
if c == '#' and stack:
stack.pop()
elif c != '#':
stack.append(c)
return "".join(stack)
return helper(S) == helper(T)