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

概述★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ ?微信公众号：为敢（WeiGanTechnologies） ?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/s

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
?微信公众号：为敢（WeiGanTechnologIEs）
?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/11371954.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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 <= words.length <= 1000 1 <= words[i].length,chars.length <= 100 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",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",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 <= words.length <= 1000 1 <= words[i].length,chars.length <= 100 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{$@H_301_132@0.ascii} 5         for c in arrChars 6         { 7             A[c - @H_301_132@97,default:@H_301_132@0] += @H_301_132@1           8         } 9         var ret:Int = @H_301_132@010         for s in words11         {12             var cnt:[Int:Int] = [Int:Int]()13             let arrS:[Int] = Array(s).map{$@H_301_132@0.ascii}14             for c in arrS15             {16                 cnt[c - @H_301_132@97,default:@H_301_132@0] += @H_301_132@1   17             }18             var found:Bool = false19             for k in @H_301_132@0..<@H_301_132@2620             {21                 if cnt[k,default:@H_301_132@0] > A[k,default:@H_301_132@0]22                 {23                     found = true24                 }25             }26             if !found27             {28                 ret += s.count29             }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 ?? @H_301_132@0)42        }       43     }44 }

总结

以上是内存溢出为你收集整理的[Swift]LeetCode1160. 拼写单词 | Find Words That Can Be Formed by Characters全部内容，希望文章能够帮你解决[Swift]LeetCode1160. 拼写单词 | Find Words That Can Be Formed by Characters所遇到的程序开发问题。

如果觉得内存溢出网站内容还不错，欢迎将内存溢出网站推荐给程序员好友。