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) O ( N ) O(N) O(N) O ( N ) O(N) O(N)104ms (35.00%)Ans 2 (Python) O ( N 2 ) O(N^2) O(N2) O ( N ) O(N) O(N)56ms (99.29%)Ans 3 (Python)– O ( N ) O(N) O(N)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