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:
Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]Return 16The two words can be "abcw", "xtfn".
Example 2:
Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]Return 4The two words can be "ab", "cd".
Example 3:
Given ["a", "aa", "aaa", "aaaa"]Return 0No such pair of words.
Credits:Special thanks to @dietpepsi for adding this problem and creating all test cases.
public class Solution { public int maxProduct(String[] words) { int max = 0; int[] masks = new int[words.length]; // alphabet masks for(int i = 0; i < masks.length; i++){ for(char c: words[i].toCharArray()){ masks[i] |= 1 << (c - 'a'); } } for(int i = 0; i < masks.length; i++){ for(int j = i + 1; j < masks.length; j++){ if((masks[i] & masks[j]) == 0){ max = Math.max(max, words[i].length() * words[j].length()); } } } return max; } }reference：
https://discuss.leetcode.com/topic/31769/32ms-java-ac-solution/17
转载于:https://www.cnblogs.com/hygeia/p/5775278.html
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