[LeetCode&Python] Problem 844. Backspace String Compare-优快云博客

本文介绍了一种在存在退格字符('#')的情况下，比较两个字符串是否相等的算法。通过实例展示了如何处理退格操作，并提供了Python代码实现。算法遵循O(N)时间复杂度和O(1)空间复杂度的要求。

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 <= 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?

class Solution(object):
    def backspaceCompare(self, S, T):
        """
        :type S: str
        :type T: str
        :rtype: bool
        """
        Sr=[]
        i=0
        while i<len(S):
            if S[i]=='#':
                if Sr:
                    Sr.pop()
            else:
                Sr.append(S[i])
            i+=1
        
        Tr=[]
        i=0
        while i<len(T):
            if T[i]=='#':
                if Tr:
                    Tr.pop()
            else:
                Tr.append(T[i])
            i+=1
        
        return Tr==Sr

转载于:https://www.cnblogs.com/chiyeung/p/10135441.html