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

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.

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)