LeetCode 392. Is Subsequence

LeetCode 392. Is Subsequence


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?

Special thanks to @pbrother for adding this problem and creating all test cases.


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)。


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?



  • 依次检查 s 中的每个字符是否出现在 t 中,并且要求 s 中后面的字符的在 t 中出现的位置对应递增。
  • 当有很多个 s ,只有一 t 个时,可以考虑用字典对 t 建立索引。键为 t 中的字符,值为 t 对应字符出现过的所有索引(递增)。
  • 查询 s 中的字符时,使用二分搜索,要求 s 中后一个字符的在 t 中的索引大于前一个字符的索引。
# -*- coding: utf-8 -*-
# @Author:             何睿
# @Create Date:        2019-08-17 13:53:03
# @Last Modified by:   何睿
# @Last Modified time: 2019-08-17 14:34:16

import bisect
from collections import defaultdict

class Solution:
    def __init__(self):
        self.__isinit = False
        self.__dict = defaultdict(list)

    def __build(self, t):
        for index, char in enumerate(t):
        self.__isinit = True

    def isSubsequence(self, s: str, t: str) -> bool:
        if not self.__isinit:

        next_ = -1
        for char in s:
            next_ = self.__check(char, next_)
            if next_ == -1:
                return False

        return True

    def __check(self, char, index):
        if char not in self.__dict:
            return -1
        next_ = bisect.bisect_right(self.__dict[char], index)
        return self.__dict[char][next_] if next_ < len(self.__dict[char]) else -1

源代码文件在 这里 。

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