[Swift]LeetCode318. 最大单词长度乘积 | Maximum Product of Wo

发布时间：2020-12-14 05:06:18 所属栏目：百科来源：网络整理

导读：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,retu

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.

Example 1:

Input: 
Output: The two words can be .["abcw","baz","foo","bar","xtfn","abcdef"]16 
Explanation:"abcw","xtfn"

Example 2:

Input: 
Output: The two words can be .["a","ab","abc","d","cd","bcd","abcd"]4 
Explanation:"ab","cd"

Example 3:

Input: 
Output: No such pair of words.["a","aa","aaa","aaaa"]0 
Explanation:

给定一个字符串数组?words，找到?length(word[i]) * length(word[j])?的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。

示例?1:

输入: 
输出: 。["abcw","abcdef"]16 
解释: 这两个单词为"abcw","xtfn"

示例 2:

输入: 
输出: 这两个单词为 。["a","abcd"]4 
解释:"ab","cd"

示例 3:

输入: 
输出: ["a","aaaa"]0 
解释: 不存在这样的两个单词。

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 }

804ms

 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 }

（编辑：李大同）

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