17. Letter Combinations of a Phone Number

Medium

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

1o-o_ 2abc 3def

4ghi_ 5jkl 6mno

7pqrs 8tuv 9wxyz

*+___ 0___ #↑__

除了0的第一個“”以外的位置的“”是為了對齊

Example:

Input: "23"

Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

大意是指要把輸入的數字按照手機鍵盤的對映關係轉成可能的字串

• 不要有重複的
• 不用在意字串順序（輸出結果陣列的排序）

Note:

雖然題目中給出輸入數字在[2, 9]，實測(2019-01-05) 0和1也有對應的對映關係

0 => " "
1 => "*"
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基本分析：

• 要求長度為n的結果，首先要求長度為n-1的結果！
• n == 1 時候結果是已知的

        "0" : [" "],
        "1" : ["*"],
        "2" : ["a", "b", "c"],
        "3" : ["d", "e", "f"],
        "4" : ["g", "h", "i"],
        "5" : ["j", "k", "l"],
        "6" : ["m", "n", "o"],
        "7" : ["p", "q", "r", "s"],
        "8" : ["t", "u", "v"],
        "9" : ["w", "x", "y", "z"]
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• 為了節約時間要有快取

基本思路

---

func letterCombinations(s: String) -> [String] {
  if s 為空字串， return 空陣列
    if cache 不包含 s {
      cache[s] = cache[s的第一個字元] * letterCombinations(s除了第一字元以外剩下的部分的結果)
    }
  return cache[s]
}
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---

程式碼

// Swift Code
class Solution {
    var cache: [String : [String]] = [
        "0" : [" "],
        "1" : ["*"],
        "2" : ["a", "b", "c"],
        "3" : ["d", "e", "f"],
        "4" : ["g", "h", "i"],
        "5" : ["j", "k", "l"],
        "6" : ["m", "n", "o"],
        "7" : ["p", "q", "r", "s"],
        "8" : ["t", "u", "v"],
        "9" : ["w", "x", "y", "z"]
    ]

    func letterCombinations(_ digits: String) -> [String] {
        if digits.isEmpty { return [] }
        if !cache.keys.contains(digits) {
            cache[digits] = letterCombinations(String(digits.prefix(1))).flatMap({ (s) -> [String] in
                letterCombinations(String(digits.suffix(digits.count - 1))).map { s + $0 }
            })
        }
        return cache[digits]!
    }
}
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TestCase

• ""
• "23"
• "203"
• "213"