給兩個(gè)list，list元素是一些str，比較兩列l(wèi)ist各自合成一列string之后是不是完全一樣（一開始沒仔細(xì)看題，原來這么簡單，考慮太多還WA一次）
思路一：兩個(gè)游標(biāo)，分別掃兩個(gè)list，中途遇到不一樣的字符就return False
1 #1662
2 #Runtime: 52 ms
3 #Memory Usage: 13.7 MB
4
5 class Solution(object):
6 def arrayStringsAreEqual(self, word1, word2):
7 """
8 :type word1: List[str]
9 :type word2: List[str]
10 :rtype: bool
11 """
12 p1 = 0
13 p2 = 0
14 pi2 = 0
15 pi1 = 0
16 while p1 < len(word1) and p2 < len(word2):
17 if pi2 < len(word2[p2]) and pi1 < len(word1[p1]) and word1[p1][pi1] != word2[p2][pi2]:
18 return False
19 pi2 += 1
20 if pi2 == len(word2[p2]):
21 p2 += 1
22 pi2 = 0
23 pi1 += 1
24 if pi1 == len(word1[p1]):
25 p1 += 1
26 pi1 = 0
27 if p1 < len(word1) or p2 < len(word2):
28 return False
29 return True
思路二：直接連起來list所有str，直接比較（看Discussion學(xué)到的簡潔寫法）
1 #1662
2 #Runtime: 22 ms
3 #Memory Usage: 13.9 MB
4
5 class Solution(object):
6 def arrayStringsAreEqual(self, word1, word2):
7 """
8 :type word1: List[str]
9 :type word2: List[str]
10 :rtype: bool
11 """
12 return ''.join(word1) == ''.join(word2)