392.Is Subsequence

Tags: Easy Binary Search Hash Table

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

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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.

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 class Solution {
public:
    bool isSubsequence(string s, string t) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int pre = -1, n = t.size();
        unordered_map<char, vector<int>> char2pos;
        for (int i = 0; i < n; ++i) char2pos[t[i]].push_back(i);
        for (char c : s) {
            auto it = upper_bound(char2pos[c].begin(), char2pos[c].end(), pre);
            if (it == char2pos[c].end()) return false;
            pre = *it;
        }
        return true;
    }
};

这道题目的扩展写法，如果是多个子串怎么办。因为子串是否存在既和字符是否存在有关，还和位置信息有关，所以用hash table来对字符和其位置关系进行映射，对于对应字符，用pre记录这个字符串上一次查找的位置，查找的时候查找第一个大于目标值的数。