快慢指针-2--LC844.比较含退格的字符串

class Solution(object):
    def backspaceCompare(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        # 1.用栈清除元素后比较
        def fun(str_):
            result = list()
            for i in str_:
                if i != '#':
                    result.append(i)
                elif result:
                    result.pop()
            return "".join(result)
        return fun(s) == fun(t)

class Solution(object):
    def backspaceCompare(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        # 2.双指针法
        # 因为#会去掉前面的字符，使用两个指针从后往前遍历，对比，不相等就返回false
        i, j = len(s)-1, len(t)-1
        # skipS, skipT是用来记录#的
        skipS, skipT = 0, 0
        while i >= 0 or j >= 0:
            # 循环字符串s
            while i >= 0:
                if s[i] == '#':
                    skipS += 1
                    i -= 1
                elif skipS:
                    skipS -= 1
                    i -= 1
                else:
                    break
            # 循环字符串t
            while j >= 0:
                if t[j] == '#':
                    skipT += 1
                    j -= 1
                elif skipT:
                    skipT -= 1
                    j -= 1
                else:
                    break
            if i>=0 and j>=0:
                if s[i] != t[j]:
                    return False
            elif i>=0 or j>=0:
                return False
            i -= 1
            j -= 1
        return True