LeetCode 792. Number of Matching Subsequences

原题链接在这里：https://leetcode.com/problems/number-of-matching-subsequences/

题目：

Given a string s and an array of strings words, return the number of words[i] that is a subsequence of s.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

• For example, "ace" is a subsequence of "abcde".

Example 1:

Input: s = "abcde", words = ["a","bb","acd","ace"]
Output: 3
Explanation: There are three strings in words that are a subsequence of s: "a", "acd", "ace".

Example 2:

Input: s = "dsahjpjauf", words = ["ahjpjau","ja","ahbwzgqnuk","tnmlanowax"]
Output: 2

Constraints:

• 1 <= s.length <= 5 * 104
• 1 <= words.length <= 5000
• 1 <= words[i].length <= 50
• s and words[i] consist of only lowercase English letters.

题解：

For each word in words, its first char of each word is the waiting char.

When iterate s, the currrent char c release corresponding waiting group in words, remove it and see the next waiting char.

If there is none, then we find a subsequence, res++.

To make sure it is liner, we use LinkedList<Character> to represent String.

Time Complexity: O(m*n + len). m = words.length. n is word length. len = s.length().

Space: O(m*n + len).

AC Java:

class Solution {
    public int numMatchingSubseq(String s, String[] words) {
        int res = 0;
        List<LinkedList<Character>> [] waiting = new List[26];
        for(int i = 0; i < 26; i++){
            waiting[i] = new ArrayList<LinkedList<Character>>();
        }
        
        for(String w : words){
            LinkedList<Character> que = new LinkedList<>();
            for(char letter : w.toCharArray()){
                que.add(letter);
            }
            
            waiting[w.charAt(0) - 'a'].add(que);
        }
        
        for(char c : s.toCharArray()){
            List<LinkedList<Character>> cur = waiting[c - 'a'];
            waiting[c - 'a'] = new ArrayList<LinkedList<Character>>();
            for(LinkedList<Character> item : cur){
                item.pollFirst();
                if(item.size() == 0){
                    res++;
                }else{
                    waiting[item.peekFirst() - 'a'].add(item);
                }
            }
        }
        
        return res;
    }
}