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题目:
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 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.
链接： https://leetcode.com/problems/maximum-product-of-word-lengths/
题解：
给定一个String array，求出不含相同字符的两个单词长度乘积的最大值。
我们可以把这道题目分为几个步骤。
这里题目给出单词只包含小写字母，所以我们还可以进一步优化空间复杂度。 另外，预先对数组按长度进行排序的话应该可以剪枝掉不少计算。 留给二刷了。
Java:
Time Complexity - O(n ^2)， Space Complexity - O(n)
public class Solution { public int maxProduct(String[] words) { if (words == null || words.length < 2) { return 0; } int max = 0; Map<String, Set<Character>> map = new HashMap<>(); for (int i = 0; i < words.length; i++) { addToMap(words[i], map); for (int j = i + 1; j < words.length; j++) { addToMap(words[j], map); if (!hasSameChar(words[i], words[j], map)) { max = Math.max(max, words[i].length() * words[j].length()); } } } return max; } private void addToMap(String word, Map<String, Set<Character>> map) { if (!map.containsKey(word)) { Set<Character> set = new HashSet<>(); for (int i = 0; i < word.length(); i++) { set.add(word.charAt(i)); } map.put(word, set); } } private boolean hasSameChar(String word1, String word2, Map<String, Set<Character>> map) { Set<Character> set1 = map.get(word1); Set<Character> set2 = map.get(word2); for (Character c : set1) { if (set2.contains(c)) { return true; } } return false; } }
Reference:
https://leetcode.com/discuss/74589/32ms-java-ac-solution
318. Maximum Product of Word Lengths
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原文地址：http://www.cnblogs.com/yrbbest/p/5300341.html
