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"""
LeetCode: 844 比较含退格的字符串
难度: Easy
链接: https://leetcode.cn/problems/backspace-string-compare/
标签: 字符串, 栈
掌握程度: ✅
解题思路: 使用栈结构遍历字符串
关联题目: 1047 删除字符串所有相邻重复项
易错点:
- #是指令符号,绝对不能存入栈中,不要把#当成普通字符添加
- 遍历完成后,通过join将两个栈转为最终成型的字符串,对比两个结果是否完全一致
PS:
- 可用指针优化算法,之后练习
"""
class Solution:
def backspaceCompare(self, s: str, t: str) -> bool:
stack_s = []
stack_t = []
for i in s:
if i == '#':
if stack_s:
stack_s.pop()
else:
stack_s.append(i)
for j in t:
if j == '#':
if stack_t:
stack_t.pop()
else:
stack_t.append(j)
res_s = ''.join(stack_s)
res_t = ''.join(stack_t)
return res_s == res_t
if __name__ == "__main__":
sol = Solution()
# 测试用例1:题目示例1,处理后均为ac
assert sol.backspaceCompare("ab#c", "ad#c") == True
# 测试用例2:题目示例2,两个字符串全部退格后均为空串
assert sol.backspaceCompare("ab##", "c#d#") == True
# 测试用例3:题目示例3,处理结果不同
assert sol.backspaceCompare("a#c", "b") == False
# 测试用例4:开头连续退格,空文本输入#无效果
assert sol.backspaceCompare("###abc", "#abc") == True
# 测试用例5:末尾退格、长度不一致的场景
assert sol.backspaceCompare("a##b", "b") == True
# 测试用例6:完全全退格清空
assert sol.backspaceCompare("####", "") == True
print("所有测试通过!")