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。

class Solution(object):
    def isLongPressedName(self, name, typed):
        """
        :type name: str
        :type typed: str
        :rtype: bool
        """
        # 下标
        i = 0
        j = 0
        while i < len(name) - 1 and j < len(typed) - 1:
            if name[i] == typed[j]:
                i += 1
                j += 1
            else:
                j += 1
        # i到末尾，j未到，返回true
        if i == len(name) - 1 and j < len(typed) - 1:
            return True
        # i、j都到末尾，且倒数第一个字符相同，返回true
        elif i == len(name) - 1 and j == len(typed) - 1 and name[-1:] == typed[-1:]:
            return True
        # 否则，返回false
        else:
            return False

if __name__ == '__main__':
    solution = Solution()
    print(solution.isLongPressedName("leelee","lleeelee"))
    print(solution.isLongPressedName("kikcxmvzi", "kiikcxxmmvvzz"))