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.
class Solution { public int maxProduct(String[] words) { int[] checker = new int[words.length]; if (words == null || words.length == 0) { return 0; } int res = 0; for (int i = 0; i < words.length; i++) { String word = words[i]; for (int j = 0; j < word.length(); j++) { char curChar = word.charAt(j); checker[i] |= 1 << curChar - ‘a‘; } } for (int i = 0; i < words.length - 1; i++) { for (int j = i + 1; j < words.length; j++) { if ((checker[i] & checker[j]) == 0) { res = Math.max(res, words[i].length() * words[j].length()); } } } return res; } }
[LC] 318. Maximum Product of Word Lengths
原文：https://www.cnblogs.com/xuanlu/p/11874918.html