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
最后修改日期: 2021年2月17日

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