925. 长按键入
思路:
i、j指针分别遍历name和typed:
若typed[j] == name[i]:i、j同步后移;
若typed[j] != name[i]:i不动、j后移;
j比i先遍历完,则返回false;否则返回true。
1 class Solution(object): 2 def isLongPressedName(self, name, typed): 3 """ 4 :type name: str 5 :type typed: str 6 :rtype: bool 7 """ 8 # 下标 9 i = 0 10 j = 0 11 while i < len(name) - 1 and j < len(typed) - 1: 12 if name[i] == typed[j]: 13 i += 1 14 j += 1 15 else: 16 j += 1 17 # i到末尾,j未到,返回true 18 if i == len(name) - 1 and j < len(typed) - 1: 19 return True 20 # i、j都到末尾,且倒数第一个字符相同,返回true 21 elif i == len(name) - 1 and j == len(typed) - 1 and name[-1:] == typed[-1:]: 22 return True 23 # 否则,返回false 24 else: 25 return False 26 27 if __name__ == '__main__': 28 solution = Solution() 29 print(solution.isLongPressedName("leelee","lleeelee")) 30 print(solution.isLongPressedName("kikcxmvzi", "kiikcxxmmvvzz"))