[Swift]LeetCode1160. 拼写单词 | Find Words That Can Be Formed by Characters

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You are given an array of strings words and a string chars.

A string is good if it can be formed by characters from chars (each character can only be used once).

Return the sum of lengths of all good strings in words.

Example 1:

Input: words = ["cat","bt","hat","tree"], chars = "atach" Output: 6 Explanation: The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6. 

Example 2:

Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr" Output: 10 Explanation: The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.

Note:

1. 1 <= words.length <= 1000
2. 1 <= words[i].length, chars.length <= 100
3. All strings contain lowercase English letters only.

---

给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。

假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。

注意：每次拼写时，chars 中的每个字母都只能用一次。

返回词汇表 words 中你掌握的所有单词的 长度之和。

示例 1：

输入：words = ["cat","bt","hat","tree"], chars = "atach"
输出：6
解释： 
可以形成字符串 "cat" 和 "hat"，所以答案是 3 + 3 = 6。

示例 2：

输入：words = ["hello","world","leetcode"], chars = "welldonehoneyr"
输出：10
解释：
可以形成字符串 "hello" 和 "world"，所以答案是 5 + 5 = 10。

提示：

1. 1 <= words.length <= 1000
2. 1 <= words[i].length, chars.length <= 100
3. 所有字符串中都仅包含小写英文字母

---

Runtime: 920 ms Memory Usage: 21 MB

 1 class Solution {
 2     func countCharacters(_ words: [String], _ chars: String) -> Int {
 3         var A:[Int:Int] = [Int:Int]()
 4         let arrChars:[Int] = Array(chars).map{$0.ascii}
 5         for c in arrChars
 6         {
 7             A[c - 97,default:0] += 1          
 8         }
 9         var ret:Int = 0
10         for s in words
11         {
12             var cnt:[Int:Int] = [Int:Int]()
13             let arrS:[Int] = Array(s).map{$0.ascii}
14             for c in arrS
15             {
16                 cnt[c - 97,default:0] += 1   
17             }
18             var found:Bool = false
19             for k in 0..<26
20             {
21                 if cnt[k,default:0] > A[k,default:0]
22                 {
23                     found = true
24                 }
25             }
26             if !found
27             {
28                 ret += s.count
29             }
30         }
31         return ret        
32     }
33 }
34 
35 //Character扩展 
36 extension Character  
37 {  
38   //Character转ASCII整数值(定义小写为整数值)
39    var ascii: Int {
40        get {
41            return Int(self.unicodeScalars.first?.value ?? 0)
42        }       
43     }
44 }