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

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].

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]。

Memory Usage: 20 MB

 1 class Solution {
 2     func numMatchingSubseq(_ S: String, _ words: [String]) -> Int {
 3         let arrS:[Character] = Array(S)
 4         var res:Int = 0
 5         var n:Int = S.count
 6         var pass:Set<String> = Set<String>()
 7         var out:Set<String> = Set<String>()
 8         for word in words
 9         {
10             let arrW:[Character] = Array(word)
11             if pass.contains(word) || out.contains(word)
12             {
13                 if pass.contains(word) {res += 1}
14                 continue
15             }
16             var i:Int = 0
17             var j:Int = 0
18             var m:Int = word.count
19             while (i < n && j < m)
20             {
21                 if arrW[j] == arrS[i] {j += 1}
22                 i += 1
23             }
24             if j == m
25             {
26                 res += 1
27                 pass.insert(word)
28             }
29             else
30             {
31                 out.insert(word)
32             }
33         }
34         return res
35     }
36 }

 1 class Solution {
 2     func numMatchingSubseq(_ S: String, _ words: [String]) -> Int {
 3         var indices = [Character: [Int]]()
 4         let S = Array(S.characters)
 5         for i in 0..<S.count {
 6             indices[S[i], default:[]].append(i)
 7         }
 8
 9         func binarySearch(_ ch: Character, _ from: Int) -> Int {
10             guard let arr = indices[ch] else { return -2 }
11             if from > arr.last! { return -2 }
12
13             var l = 0, r = arr.count - 1
14             while l < r {
15                 let mid = (l + r) / 2
16                 if arr[mid] < from {
17                     l = mid + 1
18                 } else {
19                     r = mid
20                 }
21             }
22             return arr[r]
23         }
24
25         var res = 0
26         for w in words {
27             var from = 0
28             for ch in w.characters {
29                 from = binarySearch(ch, from) + 1
30                 if from < 0 { break }
31             }
32             if from >= 0 { res += 1 }
33         }
34         return res
35     }
36 }

原文地址：https://www.cnblogs.com/strengthen/p/10547036.html