LeetCode | Is Subsequence

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?

Solution 1 :
最简单的想法是遍历s和t，依次比较后返回结果，时间复杂度为O(t.size()).

// O(t.size())bool isSubsequence(string s, string t) {    int i = 0, j = 0;     while(i < s.size() && j < t.size()){        if (s[i] == t[j])            i++;        j++;     }     return (i == s.size());}

Solution 2 :
解法一的复杂度随着t线性增长，所以试着使用string的find函数求解。STL没有给出find的实现细节，也没有提供确定的时间复杂度，但在测试的时候运行结果确实要比解法一好很多，推测find查找单个字符的平均时间复杂度可能接近常数。[仅为个人理解，如有错误还请指正]

bool isSubsequence(string s, string t) {    if (s.size() == 0)        return true;    size_t index = t.find(s[0]);    for (int i = 1; i < s.size() && index != std::string::npos; ++i){        index = t.find(s[i], index+1);    }    return (index != std::string::npos)}

最后的follow up不是很理解，如果是输入变成string数组的话就用循环咯。