## Description

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. 1 <= S.length <= 200
2. 1 <= T.length <= 200
3. S and T only contain lowercase letters and '#' characters.

• Can you solve it in O(N) time and O(1) space?

## Solutions

找到两个字符在遇到 # 就回退一个的操作后是否相同。

### 1. Stack

# Time: O(n)
# Space: O(n)
class Solution:
def backspaceCompare(self, S: str, T: str) -> bool:
s_stack, t_stack = [], []
for s in S:
if s == '#':
if s_stack:
s_stack.pop()
else:
s_stack.append(s)
for t in T:
if t == '#':
if t_stack:
t_stack.pop()
else:
t_stack.append(t)
return s_stack == t_stack

# 104/104 cases passed (32 ms)
# Your runtime beats 37.24 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)