[Swift]LeetCode1181. 先后拼接 ｜ Before and After Puzzle

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Given a list of phrases, generate a list of Before and After puzzles.

A phrase is a string that consists of lowercase English letters and spaces only. No space appears in the start or the end of a phrase. There are no consecutive spaces in a phrase.

Before and After puzzles are phrases that are formed by merging two phrases where the last word of the first phrase is the same as the first word of the second phrase.

Return the Before and After puzzles that can be formed by every two phrases phrases[i] and phrases[j] where i != j. Note that the order of matching two phrases matters, we want to consider both orders.

You should return a list of distinct strings sorted lexicographically.

 

Example 1:

Input: phrases = ["writing code","code rocks"]
Output: ["writing code rocks"]

Example 2:

Input: phrases = ["mission statement",
                  "a quick bite to eat",
                  "a chip off the old block",
                  "chocolate bar",
                  "mission impossible",
                  "a man on a mission",
                  "block party",
                  "eat my words",
                  "bar of soap"]
Output: ["a chip off the old block party",
         "a man on a mission impossible",
         "a man on a mission statement",
         "a quick bite to eat my words",
         "chocolate bar of soap"]

Example 3:

Input: phrases = ["a","b","a"]
Output: ["a"]

 

Constraints:

• 1 <= phrases.length <= 100
• 1 <= phrases[i].length <= 100

---

 

给你一个「短语」列表 phrases，请你帮忙按规则生成拼接后的「新短语」列表。

「短语」（phrase）是仅由小写英文字母和空格组成的字符串。「短语」的开头和结尾都不会出现空格，「短语」中的空格不会连续出现。

「先后拼接」（Before and After puzzles）是合并两个「短语」造成「新短语」的方法。咱们规定拼接时，第一个短语的最后一个单词 和 第二个短语的第一个单词 必须相同。

返回每两个「短语」 phrases[i] 和 phrases[j]（i != j）进行「先后拼接」获得的「新短语」。

注意，两个「短语」拼接时的顺序也很重要，咱们须要同时考虑这两个「短语」。另外，同一个「短语」能够屡次参与拼接，但「新短语」不能再参与拼接。

请你按字典序排列并返回「新短语」列表，列表中的字符串应该是 不重复的 。

 

示例 1：

输入：phrases = ["writing code","code rocks"]
输出：["writing code rocks"]

示例 2：

输入：phrases = ["mission statement",
                "a quick bite to eat",
                "a chip off the old block",
                "chocolate bar",
                "mission impossible",
                "a man on a mission",
                "block party",
                "eat my words",
                "bar of soap"]
输出：["a chip off the old block party",
      "a man on a mission impossible",
      "a man on a mission statement",
      "a quick bite to eat my words",
      "chocolate bar of soap"]

示例 3：

输入：phrases = ["a","b","a"]
输出：["a"]

 

提示：

• 1 <= phrases.length <= 100
• 1 <= phrases[i].length <= 100

---

Runtime: 112 ms

Memory Usage: 23.5 MB

 1 class Solution {
 2     func beforeAndAfterPuzzles(_ phrases: [String]) -> [String] {
 3         var map:[String:[Int]] = [String:[Int]]()
 4         var i:Int = 0
 5         for str in phrases
 6         {
 7             let first:String = str.components(separatedBy:" ").first!
 8             map[first,default:[Int]()].append(i)
 9             i += 1
10         }
11         i = 0
12         var res:Set<String> = Set<String>()
13         for str in phrases
14         {
15             let array:[String] = str.components(separatedBy:" ")
16             let last = array.last!
17             if map[last] != nil
18             {
19                 let arr:[Int] = map[last]!
20                 for index in arr
21                 {
22                     if index == i {continue}
23                     res.insert(str + phrases[index].subString(last.count))
24                 }
25             }
26             i += 1
27         }
28         return  Array(res).sorted(by:<)
29     }
30 }
31 
32 extension String {
33     // 截取字符串：从index到结束处
34     // - Parameter index: 开始索引
35     // - Returns: 子字符串
36     func subString(_ index: Int) -> String {
37         let theIndex = self.index(self.endIndex, offsetBy: index - self.count)
38         return String(self[theIndex..<endIndex])
39     }
40 }