9/15 每日一題 (遇到#時刪除前一個字元, 檢查兩組字串是否相等)

Given two strings s and t, return true 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 = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".

 

Constraints:

• 1 <= s.length, 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?

 class Solution:

    def backspaceCompare(self, s: str, t: str) -> bool:

        stack=[]

        stack2=[]

        for i in s:

            if i !='#':

                stack.append(i)

            else:

                if len(stack)>0:

                    stack.pop()

        for i in t:

            if i !='#':

                stack2.append(i)

            else:

                if len(stack2)>0:

                    stack2.pop()

        print(f'stack={stack}')

        print(f'stack2={stack2}')

        return stack == stack2