392. Is Subsequence*

https://leetcode.com/problems/is-subsequence/

题目描述

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.

C++ 实现 1

将 t 中的每个相同字符的索引保存到依次保存到同一个 vector 中, 这样我们得到了 record. 遍历 s, 用 target 表示 s[i] 的前一个字符的索引, 然后判断当前访问的字符 s[i] 的索引 *it 是否出现在 target 之后, 这可以用 upper_bound 来查找. 这种做法可以用来解决 792. Number of Matching Subsequences

class Solution {
public:
    bool isSubsequence(string s, string t) {
        unordered_map<char, vector<int>> record;
        for (int i = 0; i < t.size(); ++ i) record[t[i]].push_back(i);
        int target = -1;
        for (int i = 0; i < s.size(); ++ i) {
            if (!record.count(s[i])) return false;
            auto it = std::upper_bound(record[s[i]].begin(), record[s[i]].end(), target);
            if (it == record[s[i]].end()) return false;
            target = *it;
        }
        return true;
    }
};

C++ 实现 2

顺序访问 s 和 t.

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if (s.empty()) return true;
        int k = 0;
        for (int i = 0; i < t.size(); ++i)
            if (t[i] == s[k])
                k ++;
        return k >= s.size();
    }
};

C++ 实现 3

来自 LeetCode Submission.

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class Solution {
public:
    bool isSubsequence(string s, string t) {
        size_t index = 0;
        for (char c: s) {
            index = t.find(c, index);
            if (index == string::npos) 
                return false;
            index++;
        }
        return true;
    }
}