LeetCode 0392. Is Subsequence判断子序列【Easy】【Python】【双指针】
Problem
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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
问题
给定字符串 s 和 t ,判断 s 是否为 t 的子序列。
你可以认为 s 和 t 中仅包含英文小写字母。字符串 t 可能会很长(长度 ~= 500,000),而 s 是个短字符串(长度 <=100)。
字符串的一个子序列是原始字符串删除一些(也可以不删除)字符而不改变剩余字符相对位置形成的新字符串。(例如,"ace"是"abcde"的一个子序列,而"aec"不是)。
示例 1:
s = "abc"
, t = "ahbgdc"
返回 true
.
示例 2:
s = "axc"
, t = "ahbgdc"
返回 false
.
后续挑战 :
如果有大量输入的 S,称作S1, S2, ... , Sk 其中 k >= 10亿,你需要依次检查它们是否为 T 的子序列。在这种情况下,你会怎样改变代码?
致谢:
特别感谢 @pbrother 添加此问题并且创建所有测试用例。
思路
双指针
i 指针作为 s 的索引,j 指针作为 t 的索引。每次 j 都要右移,如果 s[i] == t[j],i 要右移。
时间复杂度: O(min(len(s), len(t)))
空间复杂度: O(1)
Python代码
class Solution(object): def isSubsequence(self, s, t): """ :type s: str :type t: str :rtype: bool """ i, j = 0, 0 while i < len(s) and j < len(t): if s[i] == t[j]: i += 1 j += 1 return i == len(s)