题目
Given a string s and a string t, check if s is subsequence of t.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, “ace” is a subsequence of “abcde” while “aec” is not).
Follow up:
If there are lots of incoming S, say S1, S2, … , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?
解题思路
- 使用两个指针指向两个字符串，如果指到相同的字符则一起向后移动，否则只移动t的
- 将s转换成一个list，遍历t，如果与s中的第一个字母相同则pop s中的第一个字母，如果s最后为空，则是true
Follow up:
先预处理，把t的存到一个字典中，key是字母，value是这个字母所有出现的位置，然后对s中的每个字符，在t中这个字母对应的位置进行二分，这里要维护每次二分到的位置，后面的字符必须要在这个位置之后才能被使用
代码
class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
pt = -1
for c in s:
while pt < len(t) - 1:
pt += 1
if c == t[pt]:
break
else:
return False
return True
class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
q = list(q)
for c in t:
if not q: return True
if c == q[0]:
q.pop(0)
return not q
Follow up:
class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
dic = collections.defaultdict(list)
for i, c in enumerate(t):
dic[c].append(i)
cur = -1
for c in s:
if c not in dic:
return False
l = dic[c]
p = bisect.bisect_left(l, cur)
if p == len(l):
return False
cur = l[p] + 1
return True
