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- """
- 925. Long Pressed Name
- 用两个index: i, j 来遍历name, typed. 当i==j时,i++, j++. 当i != j时,如果j与j-1相等那么j++。(如果j走到了typed末尾,return False。如果j != i, return False. else i++ j++
- 当跳出while循环,如果i 没有走到name的末尾,return False。如果走到末尾,check j 在type的位置,如果不在末尾,检查剩余部分是否与当前位相等。
- """
- class Solution(object):
- def isLongPressedName(self, name, typed):
- """
- :type name: str
- :type typed: str
- :rtype: bool
- """
- if not name or not typed:
- return False
- if len(typed) < len(name):
- return False
- i = j = 0
- while i < len(name) and j < len(typed):
- if name[i] == typed[j]:
- i += 1
- j += 1
- else:
- while j < len(typed) and typed[j] == typed[j-1]:
- j += 1
- if j == len(typed):
- return False
- else:
- if typed[j] != name[i]:
- return False
- else:
- i += 1
- j += 1
- if i != len(name):
- return False
- else:
- if j != len(typed):
- while j < len(typed):
- if typed[j] != typed[j-1]:
- return False
- else:
- j += 1
- return True
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