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➤微信公衆號:山青詠芝(shanqingyongzhi)
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Given a string array words
, find the maximum value of length(word[i]) * length(word[j])
where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.git
Example 1:github
Input: Output: The two words can be .["abcw","baz","foo","bar","xtfn","abcdef"]16 Explanation:"abcw", "xtfn"
Example 2:數組
Input: Output: The two words can be .["a","ab","abc","d","cd","bcd","abcd"]4 Explanation:"ab", "cd"
Example 3:微信
Input: Output: No such pair of words.["a","aa","aaa","aaaa"]0 Explanation:
給定一個字符串數組 words
,找到 length(word[i]) * length(word[j])
的最大值,而且這兩個單詞不含有公共字母。你能夠認爲每一個單詞只包含小寫字母。若是不存在這樣的兩個單詞,返回 0。app
示例 1:spa
輸入: 輸出: 。["abcw","baz","foo","bar","xtfn","abcdef"]16 解釋: 這兩個單詞爲"abcw", "xtfn"
示例 2:code
輸入: 輸出: 這兩個單詞爲 。["a","ab","abc","d","cd","bcd","abcd"]4 解釋:"ab", "cd"
示例 3:htm
輸入: 輸出: ["a","aa","aaa","aaaa"]0 解釋: 不存在這樣的兩個單詞。
332 ms
1 class Solution { 2 func maxProduct(_ words: [String]) -> Int { 3 if words.isEmpty { 4 return 0 5 } 6 7 let products = words.map{ProductHelper($0)} 8 9 var res = 0 10 11 for i in 0..<products.count { 12 for j in i+1..<products.count { 13 let p1 = products[i] 14 let p2 = products[j] 15 if p1.characters & p2.characters == 0 { 16 res = max(p1.count * p2.count, res) 17 } 18 } 19 } 20 21 return res 22 } 23 } 24 25 class ProductHelper { 26 let count : Int 27 let characters : Int 28 init(_ s : String) { 29 count = s.count 30 let arr = s.unicodeScalars 31 var r = 0 32 for c in arr { 33 r |= 1 << Int(c.value - 97) 34 } 35 characters = r 36 } 37 }
804msblog
1 class Solution { 2 func maxProduct(_ words: [String]) -> Int { 3 let aValue = "a".unicodeScalars.first!.value 4 if words.count <= 1 { return 0 } 5 var array = [Int]() 6 for word in words { 7 var a = 0 8 for c in word.unicodeScalars { 9 a = a | (1 << (c.value - aValue)) 10 } 11 array.append(a) 12 } 13 14 var result = 0 15 for i in 0 ..< array.count - 1 { 16 for j in 1 ..< array.count { 17 if array[i] & array[j] > 0 { 18 continue 19 } 20 result = max(result, words[i].count * words[j].count) 21 } 22 } 23 return result 24 } 25 }