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).
给出两个字符串s, t，判断s是否是t的子集。t大约有500000个字符， s不超过100个字符。这里的子集的判断是允许删除一些中间的字符得到的剩余字符相同（也就是不需要连续相同的字符串）
s = “abc”, t = “ahbgdc”
Return true.
s = “axc”, t = “ahbgdc”
Return false.
这道题还是比较简单的，如果s是t的子集，那么s中的所有字符必然按照相同的顺序出现在t中，所以我们可以分别建立两个字符串的索引，每当t中出现s的当前字符，就搜索s的下一个字符，直到s的所有字符均在t中找到
class Solution(object): def isSubsequence(self, s, t): """ :type s: str :type t: str :rtype: bool """ # 如果s是空字符串，那么它是任意字符串的子集 if len(s) == 0: return True index_s, index_t = 0, 0 # 遍历t for current_t in t: # 如果s的当前元素被找到，则继续寻找下一个s if s[index_s] == current_t: index_s += 1 # 如果s的所有字符都被找到，代表s是t的子集 if index_s == len(s): return True # 如果搜索完t仍然没有找全s，代表s不是t的子集 return False新闻热点
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