[Swift]LeetCode472. 连接词 | Concatenated Words
作者:互联网
Given a list of words (without duplicates), please write a program that returns all concatenated words in the given list of words.
A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.
Example:
Input: ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"] Output: ["catsdogcats","dogcatsdog","ratcatdogcat"] Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats";
"dogcatsdog" can be concatenated by "dog", "cats" and "dog";
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".
Note:
- The number of elements of the given array will not exceed
10,000
- The length sum of elements in the given array will not exceed
600,000
. - All the input string will only include lower case letters.
- The returned elements order does not matter.
给定一个不含重复单词的列表,编写一个程序,返回给定单词列表中所有的连接词。
连接词的定义为:一个字符串完全是由至少两个给定数组中的单词组成的。
示例:
输入: ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"] 输出: ["catsdogcats","dogcatsdog","ratcatdogcat"] 解释: "catsdogcats"由"cats", "dog" 和 "cats"组成; "dogcatsdog"由"dog", "cats"和"dog"组成; "ratcatdogcat"由"rat", "cat", "dog"和"cat"组成。
说明:
- 给定数组的元素总数不超过
10000
。 - 给定数组中元素的长度总和不超过
600000
。 - 所有输入字符串只包含小写字母。
- 不需要考虑答案输出的顺序。
Runtime: 8332 ms Memory Usage: 10.2 MB
1 class Solution { 2 func findAllConcatenatedWordsInADict(_ words: [String]) -> [String] { 3 var res:[String] = [String]() 4 var dict:Set<String> = Set(words) 5 for word in words 6 { 7 var n:Int = word.count 8 if n == 0 {continue} 9 var dp:[Bool] = [Bool](repeating:false,count:n + 1) 10 dp[0] = true 11 for i in 0..<n 12 { 13 if !dp[i] {continue} 14 for j in (i + 1)...n 15 { 16 var str:String = word.subString(i, j - i) 17 if j - i < n && dict.contains(str) 18 { 19 dp[j] = true 20 } 21 } 22 if dp[n] 23 { 24 res.append(word) 25 break 26 } 27 } 28 } 29 return res 30 } 31 } 32 33 extension String { 34 // 截取字符串:指定索引和字符数 35 // - begin: 开始截取处索引 36 // - count: 截取的字符数量 37 func subString(_ begin:Int,_ count:Int) -> String { 38 let start = self.index(self.startIndex, offsetBy: max(0, begin)) 39 let end = self.index(self.startIndex, offsetBy: min(self.count, begin + count)) 40 return String(self[start..<end]) 41 } 42 }
标签:String,dog,dogcatsdog,cats,ratcatdogcat,words,连接词,LeetCode472,Swift 来源: https://www.cnblogs.com/strengthen/p/10347611.html