LeetCode | 0392. Is Subsequence判断子序列【Python】

标签：双指针   problem   potential   sum   字符   href   pre   elf   判断   

LeetCode 0392. Is Subsequence判断子序列【Easy】【Python】【双指针】

Problem

LeetCode

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

力扣

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)

代码地址

GitHub链接

LeetCode | 0392. Is Subsequence判断子序列【Python】

标签：双指针   problem   potential   sum   字符   href   pre   elf   判断   

原文地址：https://www.cnblogs.com/wonz/p/12335210.html