Description
Your friend is typing his name into a keyboard. Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times.
You examine the typed characters of the keyboard. Return True if it is possible that it was your friends name, with some characters (possibly none) being long pressed.
你的朋友正在使用键盘输入他的名字 name。偶尔，在键入字符 c 时，按键可能会被长按，而字符可能被输入 1 次或多次。
你将会检查键盘输入的字符 typed。如果它对应的可能是你的朋友的名字（其中一些字符可能被长按），那么就返回 True。
题目链接：https://leetcode.com/problems/long-pressed-name/
Difficulty: easy
Example 1:
Input: name = "alex", typed = "aaleex"
Output: true
Explanation: 'a' and 'e' in 'alex' were long pressed.
Example 2:
Input: name = "saeed", typed = "ssaaedd"
Output: false
Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
Example 3:
Input: name = "leelee", typed = "lleeelee"
Output: true
Example 4:
Input: name = "laiden", typed = "laiden"
Output: true
Explanation: It's not necessary to long press any character.
Note:
- name.length <= 1000
- typed.length <= 1000
- The characters of name and typed are lowercase letters.
分析
- 由题意知，由于长按导致的情况是某一个字母出现的次数变多，且是连续的，定义一个get用于返回连续的字母及其出现的次数，形如get(“aaleex”)->[(‘a’,2),(‘l’,1),(‘e’,2),(‘x’,1)]
- 若name和typed都为空则返回True；
- 若get(name)和get(typed)的长度不一致，返回False；
- 反之，若每一项中get(typed)的次数大于等于get(name)，则返回True，反之False。
参考代码
class Solution:
def isLongPressedName(self, name, typed):
def get(s):
if(len(s)==0):
return []
li=[]
tar=s[0]
index=0
for i in s:
if(i!=tar):
li.append((tar,index))
index=0
tar=i
else:
index+=1
return li
l1=get(name)
l2=get(typed)
if(len(l1)==0 and len(l2)==0):
return True
if(len(l1) == len(l2)):
for i in range(len(l1)):
if(l1[i][0]==l2[i][0] and l1[i][1]<=l2[i][1]):
continue
else:
return False
return True
else:
return False