[LeetCode] 844. Backspace String Compare_Easy tag: Stack **Two pointers

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?

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)

巴特西

[LeetCode] 844. Backspace String Compare_Easy tag: Stack **Two pointers

最新文章

热门文章