LeetCode题解(0925)：判断字符串是长按键入的错误(Python)

题目：原题链接（简单）

标签：字符串、双指针

解法时间复杂度空间复杂度执行用时Ans 1 (Python)$O(N+K)$$O(1)$44ms (56.90%)Ans 2 (Python)$O(N+K)$$O(N+K)$48ms (32.38%)Ans 3 (Python)

解法一（双指针）：

def isLongPressedName(self, name: str, typed: str) -> bool: idx = 0 for i in range(len(typed)): if idx < len(name) and typed[i] == name[idx]: idx += 1 else: if i == 0 or typed[i] != name[idx - 1]: return False return idx == len(name)

解法二（按块比较长度）：

def isLongPressedName(self, name: str, typed: str) -> bool: g1 = [(k, len(list(group))) for k, group in itertools.groupby(name)] g2 = [(k, len(list(group))) for k, group in itertools.groupby(typed)] if len(g1) != len(g2): return False return all(k1 == k2 and v1 <= v2 for (k1, v1), (k2, v2) in zip(g1, g2))