[Swift]LeetCode792. 匹配子序列的单词数 | Number of Matching Subsequences

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： https://www.cnblogs.com/strengthen/p/10547036.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Given string S and a dictionary of words words, find the number of words[i] that is a subsequence of S.

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

Note:

• All words in words and S will only consists of lowercase letters.
• The length of S will be in the range of [1, 50000].
• The length of words will be in the range of [1, 5000].
• The length of words[i] will be in the range of [1, 50].

---

给定字符串 S 和单词字典 words, 求 words[i] 中是 S 的子序列的单词个数。

示例:
输入: 
S = "abcde"
words = ["a", "bb", "acd", "ace"]
输出: 3
解释: 有三个是 S 的子序列的单词: "a", "acd", "ace"。

注意:

• 所有在words和 S 里的单词都只由小写字母组成。
• S 的长度在 [1, 50000]。
• words 的长度在 [1, 5000]。
• words[i]的长度在[1, 50]。

---

Runtime: 740 ms

Memory Usage: 20 MB

class Solution {
    func numMatchingSubseq(_ S: String, _ words: [String]) -> Int {
        let arrS:[Character] = Array(S)
        var res:Int = 0
        var n:Int = S.count
        var pass:Set<String> = Set<String>()
        var out:Set<String> = Set<String>()
        for word in words
        {
            let arrW:[Character] = Array(word)
            if pass.contains(word) || out.contains(word)
            {
                if pass.contains(word) {res += 1}
                continue
            }
            var i:Int = 0
            var j:Int = 0
            var m:Int = word.count
            while (i < n && j < m)
            {
                if arrW[j] == arrS[i] {j += 1}
                i += 1
            }
            if j == m 
            {
                res += 1
                pass.insert(word)
            }
            else
            {
                out.insert(word)
            }            
        }
        return res
    }
}

---

1168ms

class Solution {
    func numMatchingSubseq(_ S: String, _ words: [String]) -> Int {
        var indices = [Character: [Int]]()
        let S = Array(S.characters)
        for i in 0..<S.count {
            indices[S[i], default:[]].append(i)
        }
        
        func binarySearch(_ ch: Character, _ from: Int) -> Int {
            guard let arr = indices[ch] else { return -2 }
            if from > arr.last! { return -2 }
            
            var l = 0, r = arr.count - 1
            while l < r {
                let mid = (l + r) / 2
                if arr[mid] < from {
                    l = mid + 1
                } else {
                    r = mid
                }
            }
            return arr[r]
        }
        
        var res = 0
        for w in words {
            var from = 0
            for ch in w.characters {
                from = binarySearch(ch, from) + 1
                if from < 0 { break }
            }
            if from >= 0 { res += 1 }
        }
        return res
    }
}