392. Is Subsequence
难度: 中等
刷题内容
原题连接
- https://leetcode.com/problems/is-subsequence
内容描述
Given a string s and a string t, check if s is subsequence of t.
You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).
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).
Example 1:
s = "abc", t = "ahbgdc"
Return true.
Example 2:
s = "axc", t = "ahbgdc"
Return false.
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?
解题方案
思路 1
follow up question很有意思
最naive的思路表现形式如下：
beats 53.74%
class Solution(object):
def isSubsequence(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
if s == '': return True
for i in xrange(len(t)):
if t[i] == s[0]:
return self.isSubsequence(s[1:],t[i+1:])
return False
思路 2
递归操作以及对字符串的操作太过于昂贵，所以用index来处理，节省了时间和空间
class Solution(object):
def isSubsequence(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
if s == '': return True
ps, pt = 0, 0
while ps < len(s) and pt < len(t):
if s[ps] == t[pt]:
ps += 1
pt += 1
return ps >= len(s)
精妙绝伦！！
