Easy
i指向s的开头,另一个指向t的开头t,当遇到s[i]与当前元素相同时,i向后移动一位i移动完了之后,说明匹配成功,否则匹配失败s == ""返回True以上,AC!
python3
class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if s is "":
return True
i = 0
for l in t:
if s[i] == l:
i += 1
if i >= len(s):
return True
return False