题目:原题链接(简单)
标签:字符串、字符串-替换函数、栈、正则表达式、正则表达式-替换函数
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) |
| 104ms (35.00%) | |
Ans 2 (Python) | 56ms (99.29%) | ||
Ans 3 (Python) | – | 56ms (99.29%) |
LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一(栈):
def removeDuplicates(self, S: str) -> str:
stack = []
for s in S:
if len(stack) == 0 or stack[-1] != s:
stack.append(s)
else:
stack.pop(-1)
return "".join(stack)
解法二(替换函数):
def removeDuplicates(self, S: str) -> str:
duplicates = {2 * ch for ch in string.ascii_lowercase}
last_length = -1
while len(S) != last_length:
last_length = len(S)
for d in duplicates:
S = S.replace(d, "")
return S
解法三(正则表达式):
def removeDuplicates(self, S: str) -> str:
last_length = -1
while len(S) != last_length:
last_length = len(S)
S = re.sub(r"(.)\1", "", S)
return S