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).
給出兩個字符串s, t，判斷s是否是t的子集。t大約有500000個字符， s不超過100個字符。這里的子集的判斷是允許刪除一些中間的字符得到的剩余字符相同（也就是不需要連續相同的字符串）
s = “abc”, t = “ahbgdc”
Return true.
s = “axc”, t = “ahbgdc”
Return false.
這道題還是比較簡單的，如果s是t的子集，那么s中的所有字符必然按照相同的順序出現在t中，所以我們可以分別建立兩個字符串的索引，每當t中出現s的當前字符，就搜索s的下一個字符，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 # 遍歷t for current_t in t: # 如果s的當前元素被找到，則繼續尋找下一個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仍然沒有找全s，代表s不是t的子集 return False新聞熱點
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