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

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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.

Example 1:

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

Example 2:

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

Example 3:

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

---

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

示例 2:

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

示例 3:

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

---

332 ms

class Solution {    
    func maxProduct(_ words: [String]) -> Int {
        if words.isEmpty {
            return 0
        }
        
        let products = words.map{ProductHelper($0)}
        
        var res = 0
        
        for i in 0..<products.count {
            for j in i+1..<products.count {
                let p1 = products[i]
                let p2 = products[j]
                if p1.characters & p2.characters == 0 {
                    res = max(p1.count * p2.count, res)
                }
            }
        }
        
        return res
    }
}

class ProductHelper {
    let count : Int
    let characters : Int
    init(_ s : String) {
        count = s.count
        let arr = s.unicodeScalars
        var r = 0
        for c in arr {
            r |= 1 << Int(c.value - 97) 
        }
        characters = r
    }
}

---

804ms

class Solution {
    func maxProduct(_ words: [String]) -> Int {
        let aValue = "a".unicodeScalars.first!.value
        if words.count <= 1 { return 0 }
        var array = [Int]()
        for word in words {
            var a = 0
            for c in word.unicodeScalars {
                a = a | (1 << (c.value - aValue))
            }
            array.append(a)
        }
        
        var result = 0
        for i in 0 ..< array.count - 1 {
            for j in 1 ..< array.count {
                if array[i] & array[j] > 0 {
                    continue
                }
                result = max(result, words[i].count * words[j].count)
            }
        }
        return result
    }    
}