
**解析:**Version 1,遍历数组中的字符串,同时遍历s,如果数组中的字符不等于对应位置的字符,直接返回False,当s遍历结束时,判断数组中当前位置的字符串是否遍历完成,如果数组遍历结束,判断字符串s是否遍历完成。Version 2直接按单词比较。
class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
n = len(s)
i = 0
for word in words:
for j in range(len(word)):
if s[i] == word[j]:
i += 1
else:
return False
if i == n:
return j == len(word) - 1
return i == nclass Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
n = len(s)
i = 0
for word in words:
m = len(word)
if s[i:i+m] == word:
i += m
else:
return False
if i == n:
return True
return i == n