318-Maximum Product of Word Lengths

Description

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.

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Example 1:

Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".

Example 2:

Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".

Example 3:

Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.

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問題描述

給定字串陣列words, 返回words中沒有相同字母的兩個字串的最大的長度之積。words中的單詞都只有小寫字母。如果沒有字母都不相同的兩個字串， 返回0

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問題分析

關鍵是使用一個整數來保留字串words[i]中字母的資訊, ‘a’對應最低位為1, ‘b’對應次低位為1, 依次類推。
然後遍歷字串， 找出最大的長度之積即可

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解法

class Solution {
    public static int maxProduct(String[] words) {
        if(words == null || words.length == 0) return 0;

        int len = words.length;
        int[] value = new int[len];

        for(int i = 0; i < len; i++){
            String tmp = words[i];
            //關鍵是這裡的轉換
            for(int j = 0; j < tmp.length(); j++)   value[i] |= 1 << (tmp.charAt(j) - 'a');
        }

        int maxProduct = 0;
        for(int i = 0; i < len; i++){
            for(int j = i + 1; j < len; j++){
                //注意這裡使用按位與判斷
                if((value[i] & value[j]) == 0 && (words[i].length() * words[j].length() > maxProduct)){
                    maxProduct = words[i].length() * words[j].length();
                }
            }
        }

        return maxProduct;
    }
}